#set document(title: "6.4 Dirac Delta and Impulse Response", author: "Jiří Lebl") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 6.4#h(0.6em)Dirac Delta and Impulse Response === Rectangular Pulse Often in applications we study a physical system by putting in a short pulse and then seeing what the system does. The resulting behavior is often called #emph[impulse response]. Let us see what we mean by a pulse. The simplest kind of a pulse is a simple rectangular pulse defined by #math.equation(block: true, alt: "0 i f t less than a ,; M i f a less than or equal to t less than b ,; 0 i f b less than or equal to t .")[$0 upright(i) upright(f) " " " " " " " " " " t < a , \ M upright(i) upright(f) " " a ≤ t < b , \ 0 upright(i) upright(f) " " b ≤ t .$] Notice that #math.equation(block: true, alt: "φ open parenthesis t close parenthesis equals M open parenthesis u open parenthesis t minus a close parenthesis minus u open parenthesis t minus b close parenthesis close parenthesis ,")[$φ ( t ) = M ( u ( t − a ) − u ( t − b ) ) ,$] where #math.equation(block: false, alt: "u open parenthesis t close parenthesis")[$u ( t )$] is the unit step function (see Figure #math.equation(block: false, alt: "1")[$1$] for a graph). #figure(figph[Graph that is on the x-axis from (0,0) to (0,0.5), goes up to (0,2), then horizontal to (1,2) then down to the x-axis again.], alt: "Graph that is on the x-axis from (0,0) to (0,0.5), goes up to (0,2), then horizontal to (1,2) then down to the x-axis again.", caption: [Figure #math.equation(block: false, alt: "1")[$1$]: Sample square pulse with #math.equation(block: false, alt: "a equals 0.5")[$a = 0.5$], #math.equation(block: false, alt: "b equals 1")[$b = 1$], and #math.equation(block: false, alt: "M equals 2")[$M = 2$].]) Let us take the Laplace transform of a square pulse, #math.equation(block: true, alt: "ℒ { φ open parenthesis t close parenthesis } equals ℒ { M open parenthesis u open parenthesis t minus a close parenthesis minus u open parenthesis t minus b close parenthesis close parenthesis } equals M the fraction e to the power minus a s minus e to the power minus b s over s .")[$ℒ \{ φ ( t ) \} = ℒ \{ M ( u ( t − a ) − u ( t − b ) ) \} = M frac(e^(− a s) − e^(− b s), s) .$] For simplicity we let #math.equation(block: false, alt: "a equals 0")[$a = 0$] and it is convenient to set #math.equation(block: false, alt: "M equals the fraction 1 over b")[$M = frac(1, b)$] to have #math.equation(block: true, alt: "∫ 0 infinity φ open parenthesis t close parenthesis d t equals 1")[$∫_(0)^(∞) φ ( t ) #h(0.278em) d t = 1$] That is, to have the pulse have “unit mass.” For such a pulse we compute #math.equation(block: true, alt: "ℒ { φ open parenthesis t close parenthesis } equals ℒ { the fraction u open parenthesis t close parenthesis minus u open parenthesis t minus b close parenthesis over b } equals the fraction 1 minus e to the power minus b s over b s .")[$ℒ \{ φ ( t ) \} = ℒ \{ frac(u ( t ) − u ( t − b ), b) \} = frac(1 − e^(− b s), b s) .$] We generally want #math.equation(block: false, alt: "b")[$b$] to be very small. That is, we wish to have the pulse be very short and very tall. By letting #math.equation(block: false, alt: "b")[$b$] go to zero we arrive at the concept of the #emph[Dirac delta function]. The book's own normalized pulse φ(t) = (u(t) − u(t−b))/b — the a = 0, M = 1/b case this section sets up right before it takes the limit — with time t on the horizontal axis. Drag b from 4 down to 0.2 and watch the rectangle go short-and-wide → tall-and-thin: the height 1/b is not chosen, it is forced by the area staying exactly 1, which is precisely the limit that defines δ(t). At b = 0.5 the rectangle is 0.5 wide and 2 tall — the same shape as the book's own sample square pulse (a = 0.5, b = 1, M = 2), slid back to start at the origin. There are no vertical sides at t = 0 and t = b: those are genuine jump discontinuities, not gaps in the drawing. {"functions":\[{"color":"\#1f4e79","domain":null,"expression":"(1/(2\*b))\*(x/abs(x) - (x-b)/abs(x-b))","expression2":"","id":"f-pulse","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true}\],"grid":{"axisColor":"\#333333","gridColor":"\#cccccc","majorSpacing":1,"minorGridColor":"\#eeeeee","showAxes":true,"showLabels":true},"parameters":\[{"id":"p-b","label":"Pulse width b","max":4,"min":0.2,"name":"b","step":0.05,"unit":"","value":0.5}\],"title":"Unit-mass pulse as b \\u2192 0","version":1,"viewport":{"centerX":2.6,"centerY":2.5,"scale":78}} === 6.4.2Delta Function The #emph[Dirac delta function]#math.equation(block: false, alt: "to the power 1")[$1$] is not exactly a function; it is sometimes called a generalized function. We avoid unnecessary details and simply say that it is an object that does not really make sense unless we integrate it. The motivation is that we would like a “function” #math.equation(block: false, alt: "δ open parenthesis t close parenthesis")[$δ ( t )$] such that for any continuous function #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] we have #math.equation(block: true, alt: "∫ minus infinity infinity δ open parenthesis t close parenthesis f open parenthesis t close parenthesis d t equals f open parenthesis 0 close parenthesis")[$∫_(− ∞)^(∞) δ ( t ) f ( t ) d t = f ( 0 )$] The formula should hold if we integrate over any interval that contains 0, not just #math.equation(block: false, alt: "open parenthesis minus infinity , infinity close parenthesis")[$( − ∞ , ∞ )$]. So #math.equation(block: false, alt: "δ open parenthesis t close parenthesis")[$δ ( t )$] is a “function” with all its “mass” at the single point #math.equation(block: false, alt: "t equals 0")[$t = 0$]. In other words, for any interval #math.equation(block: false, alt: "[ c , d ]")[$[ c , d ]$] #math.equation(block: true, alt: "1 i f t h e i n t e r v a l [ c , d ] c o n t a i n s 0 , i . e . c less than or equal to 0 less than or equal to d ,; 0 o t h e r w i s e .")[$1 upright(i) upright(f) " " upright(t) upright(h) upright(e) " " upright(i) upright(n) upright(t) upright(e) upright(r) upright(v) upright(a) upright(l) " " [ c , d ] " " upright(c) upright(o) upright(n) upright(t) upright(a) upright(i) upright(n) upright(s) " " 0 , " " upright(i) . upright(e) . " " c ≤ 0 ≤ d , \ 0 upright(o) upright(t) upright(h) upright(e) upright(r) upright(w) upright(i) upright(s) upright(e) .$] Unfortunately there is no such function in the classical sense. You could informally think that #math.equation(block: false, alt: "δ open parenthesis t close parenthesis")[$δ ( t )$] is zero for #math.equation(block: false, alt: "t not equal to 0")[$t ≠ 0$] and somehow infinite at #math.equation(block: false, alt: "t equals 0")[$t = 0$]. A good way to think about #math.equation(block: false, alt: "δ open parenthesis t close parenthesis")[$δ ( t )$] is as a limit of short pulses whose integral is 1. For example, suppose that we have a square pulse #math.equation(block: false, alt: "φ open parenthesis t close parenthesis")[$φ ( t )$] as above with #math.equation(block: false, alt: "a equals 0")[$a = 0$], #math.equation(block: false, alt: "M equals the fraction 1 over b")[$M = frac(1, b)$], that is #math.equation(block: false, alt: "φ open parenthesis t close parenthesis equals the fraction u open parenthesis t close parenthesis minus u open parenthesis t minus b close parenthesis over b")[$φ ( t ) = frac(u ( t ) − u ( t − b ), b)$]. Compute #math.equation(block: true, alt: "∫ minus infinity infinity φ open parenthesis t close parenthesis f open parenthesis t close parenthesis d t equals ∫ minus infinity infinity the fraction u open parenthesis t close parenthesis minus u open parenthesis t minus b close parenthesis over b f open parenthesis t close parenthesis d t equals the fraction 1 over b ∫ 0 b f open parenthesis t close parenthesis d t .")[$∫_(− ∞)^(∞) φ ( t ) f ( t ) d t = ∫_(− ∞)^(∞) frac(u ( t ) − u ( t − b ), b) f ( t ) d t = frac(1, b) ∫_(0)^(b) f ( t ) d t .$] If #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is continuous at #math.equation(block: false, alt: "t equals 0")[$t = 0$], then for very small #math.equation(block: false, alt: "b")[$b$], the function #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is approximately equal to #math.equation(block: false, alt: "f open parenthesis 0 close parenthesis")[$f ( 0 )$] on the interval #math.equation(block: false, alt: "[ 0 , b close parenthesis")[$[ 0 , b )$]. We approximate the integral #math.equation(block: true, alt: "the fraction 1 over b ∫ 0 b f open parenthesis t close parenthesis d t approximately equals the fraction 1 over b ∫ 0 b f open parenthesis 0 close parenthesis d t equals f open parenthesis 0 close parenthesis .")[$frac(1, b) ∫_(0)^(b) f ( t ) d t ≈ frac(1, b) ∫_(0)^(b) f ( 0 ) d t = f ( 0 ) .$] Therefore, #math.equation(block: true, alt: "lim sub b → 0 ∫ minus infinity infinity φ open parenthesis t close parenthesis f open parenthesis t close parenthesis d t equals lim sub b → 0 the fraction 1 over b ∫ 0 b f open parenthesis t close parenthesis d t equals f open parenthesis 0 close parenthesis")[$lim_(b → 0) ∫_(− ∞)^(∞) φ ( t ) f ( t ) d t = lim_(b → 0) frac(1, b) ∫_(0)^(b) f ( t ) d t = f ( 0 )$] Let us therefore accept #math.equation(block: false, alt: "δ open parenthesis t close parenthesis")[$δ ( t )$] as an object that is possible to integrate. We often want to shift #math.equation(block: false, alt: "δ")[$δ$] to another point, for example #math.equation(block: false, alt: "δ open parenthesis t minus a close parenthesis")[$δ ( t − a )$]. In that case we have #math.equation(block: true, alt: "∫ minus infinity infinity δ open parenthesis t minus a close parenthesis f open parenthesis t close parenthesis d t equals f open parenthesis a close parenthesis .")[$∫_(− ∞)^(∞) δ ( t − a ) f ( t ) d t = f ( a ) .$] Note that #math.equation(block: false, alt: "δ open parenthesis a minus t close parenthesis")[$δ ( a − t )$] is the same object as #math.equation(block: false, alt: "δ open parenthesis t minus a close parenthesis")[$δ ( t − a )$]. In other words, the convolution of #math.equation(block: false, alt: "δ open parenthesis t close parenthesis")[$δ ( t )$] with #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is again #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$], #math.equation(block: true, alt: "open parenthesis f * δ close parenthesis open parenthesis t close parenthesis equals ∫ 0 t δ open parenthesis t minus s close parenthesis f open parenthesis s close parenthesis d s equals f open parenthesis t close parenthesis")[$( f * δ ) ( t ) = ∫_(0)^(t) δ ( t − s ) f ( s ) d s = f ( t )$] As we can integrate #math.equation(block: false, alt: "δ open parenthesis t close parenthesis")[$δ ( t )$], let us compute its Laplace transform. #math.equation(block: true, alt: "ℒ { δ open parenthesis t minus a close parenthesis } equals ∫ 0 infinity e to the power minus s t δ open parenthesis t minus a close parenthesis d t equals e to the power minus a s .")[$ℒ \{ δ ( t − a ) \} = ∫_(0)^(∞) e^(− s t) δ ( t − a ) d t = e^(− a s) .$] In particular, #math.equation(block: true, alt: "ℒ { δ open parenthesis t close parenthesis } equals 1 .")[$ℒ \{ δ ( t ) \} = 1 .$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Note] Notice that the Laplace transform of #math.equation(block: false, alt: "δ open parenthesis t minus a close parenthesis")[$δ ( t − a )$] looks like the Laplace transform of the derivative of the Heaviside function #math.equation(block: false, alt: "u open parenthesis t minus a close parenthesis")[$u ( t − a )$], if we could differentiate the Heaviside function. First notice #math.equation(block: true, alt: "ℒ { δ open parenthesis t minus a close parenthesis } equals the fraction e to the power minus a s over s .")[$ℒ \{ δ ( t − a ) \} = frac(e^(− a s), s) .$] To obtain what the Laplace transform of the derivative would be we multiply by #math.equation(block: false, alt: "s")[$s$], to obtain #math.equation(block: false, alt: "e to the power minus a s")[$e^(− a s)$], which is the Laplace transform of #math.equation(block: false, alt: "δ open parenthesis t minus a close parenthesis")[$δ ( t − a )$]. We see the same thing using integration, #math.equation(block: true, alt: "∫ 0 t δ open parenthesis s minus a close parenthesis d s equals u open parenthesis t minus a close parenthesis")[$∫_(0)^(t) δ ( s − a ) d s = u ( t − a )$] So in a certain sense #math.equation(block: true, alt: "the fraction d over d t [ u open parenthesis t minus a close parenthesis ] equals δ open parenthesis t minus a close parenthesis")[$frac(d, d t) [ u ( t − a ) ] = δ ( t − a )$] This line of reasoning allows us to talk about derivatives of functions with jump discontinuities. We can think of the derivative of the Heaviside function #math.equation(block: false, alt: "u open parenthesis t minus a close parenthesis")[$u ( t − a )$] as being somehow infinite at #math.equation(block: false, alt: "a")[$a$], which is precisely our intuitive understanding of the delta function. ] #examplebox("Example 1")[][ Compute #math.equation(block: true, alt: "ℒ to the power minus 1 { the fraction s plus 1 over s } .")[$ℒ^(− 1) \{ frac(s + 1, s) \} .$] So far we have always looked at proper rational functions in the #math.equation(block: false, alt: "s")[$s$] variable. That is, the numerator was always of lower degree than the denominator. Not so with #math.equation(block: false, alt: "the fraction s plus 1 over s")[$frac(s + 1, s)$]. We write, #math.equation(block: true, alt: "ℒ to the power minus 1 { the fraction s plus 1 over s } equals ℒ to the power minus 1 { 1 plus the fraction 1 over s } equals ℒ to the power minus 1 { 1 } plus ℒ to the power minus 1 { the fraction 1 over s } equals δ open parenthesis t close parenthesis plus 1 .")[$ℒ^(− 1) \{ frac(s + 1, s) \} = ℒ^(− 1) \{ 1 + frac(1, s) \} = ℒ^(− 1) \{ 1 \} + ℒ^(− 1) \{ frac(1, s) \} = δ ( t ) + 1 .$] The resulting object is a generalized function and only makes sense when put underneath an integral. ] === Impulse Response As we said before, in the differential equation #math.equation(block: false, alt: "L x equals f open parenthesis t close parenthesis")[$L x = f ( t )$], we think of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] as input, and #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$] as the output. Often it is important to find the response to an impulse, and then we use the delta function in place of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]. The solution to #math.equation(block: true, alt: "L x equals δ open parenthesis t close parenthesis")[$L x = δ ( t )$] is called the #emph[impulse response]. #examplebox("Example 2")[][ Solve (find the impulse response) #math.equation(block: true, alt: "x double prime plus ω 0 2 x equals δ open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0 .")[$x^(″) + ω_(0)^(2) x = δ ( t ) , #h(1em) x ( 0 ) = 0 , #h(1em) x^(′) ( 0 ) = 0 .$] We first apply the Laplace transform to the equation. Denote the transform of #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$] by #math.equation(block: false, alt: "X open parenthesis s close parenthesis")[$X ( s )$]. #math.equation(block: true, alt: "s squared X open parenthesis s close parenthesis plus ω 0 2 X open parenthesis s close parenthesis equals 1 , and so X open parenthesis s close parenthesis equals the fraction 1 over s squared plus ω 0 2 .")[$s^(2) X ( s ) + ω_(0)^(2) X ( s ) = 1 , #h(1em) "and so" #h(1em) X ( s ) = frac(1, s^(2) + ω_(0)^(2)) .$] Taking the inverse Laplace transform we obtain #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals the fraction sin open parenthesis ω sub 0 t close parenthesis over ω sub 0 .")[$x ( t ) = frac(sin ( ω_(0) t ), ω_(0)) .$] ] Let us notice something about the above example. In Example 6.3.4, we found that when the input was #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$], then the solution to #math.equation(block: false, alt: "L x equals f open parenthesis t close parenthesis")[$L x = f ( t )$] was given by #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals ∫ 0 t f open parenthesis τ close parenthesis the fraction sin open parenthesis ω sub 0 open parenthesis t minus τ close parenthesis close parenthesis over ω sub 0 d τ .")[$x ( t ) = ∫_(0)^(t) f ( τ ) frac(sin ( ω_(0) ( t − τ ) ), ω_(0)) d τ .$] Notice that #emph[the solution for an arbitrary input is given as convolution with the impulse response]. Let us see why. The key is to notice that for functions #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$] and #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$], #math.equation(block: true, alt: "open parenthesis x * f close parenthesis double prime open parenthesis t close parenthesis equals the fraction d squared over d t squared [ ∫ 0 t f open parenthesis τ close parenthesis x open parenthesis t minus τ close parenthesis d τ ] equals ∫ 0 t f open parenthesis τ close parenthesis x double prime open parenthesis t minus τ close parenthesis d τ equals open parenthesis x double prime * f close parenthesis open parenthesis t close parenthesis .")[$( x * f )^(″) ( t ) = frac(d^(2), d t^(2)) [ ∫_(0)^(t) f ( τ ) x ( t − τ ) d τ ] = ∫_(0)^(t) f ( τ ) x^(″) ( t − τ ) d τ = ( x^(″) * f ) ( t ) .$] We simply differentiate twice under the integral,#math.equation(block: false, alt: "squared")[$2$] the details are left as an exercise. And so if we convolve the entire equation (6.4.1), the left hand side becomes #math.equation(block: true, alt: "open parenthesis x double prime plus ω 0 2 x close parenthesis * f equals open parenthesis x double prime * f close parenthesis plus ω 0 2 open parenthesis x * f close parenthesis equals open parenthesis x * f close parenthesis double prime plus ω 0 2 open parenthesis x * f close parenthesis .")[$( x^(″) + ω_(0)^(2) x ) * f = ( x^(″) * f ) + ω_(0)^(2) ( x * f ) = ( x * f )^(″) + ω_(0)^(2) ( x * f ) .$] The right hand side becomes #math.equation(block: true, alt: "open parenthesis δ * f close parenthesis open parenthesis t close parenthesis equals f open parenthesis t close parenthesis .")[$( δ * f ) ( t ) = f ( t ) .$] Therefore #math.equation(block: false, alt: "y open parenthesis t close parenthesis equals open parenthesis x * f close parenthesis open parenthesis t close parenthesis")[$y ( t ) = ( x * f ) ( t )$] is the solution to #math.equation(block: true, alt: "y double prime plus ω 0 2 y equals f open parenthesis t close parenthesis .")[$y^(″) + ω_(0)^(2) y = f ( t ) .$] This procedure works in general for other linear equations #math.equation(block: false, alt: "L x equals f open parenthesis t close parenthesis")[$L x = f ( t )$]. If you determine the impulse response, you also know how to obtain the output #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$] for any input #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] by simply convolving the impulse response and the input #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]. === Three-Point Beam Bending Let us give another quite different example where the delta function turns up: Representing point loads on a steel beam. Consider a beam of length #math.equation(block: false, alt: "L")[$L$], resting on two simple supports at the ends. Let #math.equation(block: false, alt: "x")[$x$] denote the position on the beam, and let #math.equation(block: false, alt: "y open parenthesis x close parenthesis")[$y ( x )$] denote the deflection of the beam in the vertical direction. The deflection #math.equation(block: false, alt: "y open parenthesis x close parenthesis")[$y ( x )$] satisfies the #emph[Euler-Bernoulli equation],#math.equation(block: false, alt: "cubed")[$3$] #math.equation(block: true, alt: "E I the fraction d to the power 4 y over d x to the power 4 equals F open parenthesis x close parenthesis ,")[$E I frac(d^(4) y, d x^(4)) = F ( x ) ,$] where #math.equation(block: false, alt: "E")[$E$] and #math.equation(block: false, alt: "I")[$I$] are constants#math.equation(block: false, alt: "to the power 4")[$4$] and #math.equation(block: false, alt: "F open parenthesis x close parenthesis")[$F ( x )$] is the force applied per unit length at position #math.equation(block: false, alt: "x")[$x$]. The situation we are interested in is when the force is applied at a single point as in Figure #math.equation(block: false, alt: "2")[$2$]. #figure(figph[Picture of a bar that is pushed up on the ends and down in the center.], alt: "Picture of a bar that is pushed up on the ends and down in the center.", caption: [Figure #math.equation(block: false, alt: "2")[$2$]: Three-point bending.]) In this case the equation becomes #math.equation(block: true, alt: "E I the fraction d to the power 4 y over d x to the power 4 equals minus F δ open parenthesis x minus a close parenthesis")[$E I frac(d^(4) y, d x^(4)) = − F δ ( x − a )$] where #math.equation(block: false, alt: "x equals a")[$x = a$] is the point where the mass is applied. #math.equation(block: false, alt: "F")[$F$] is the force applied and the minus sign indicates that the force is downward, that is, in the negative #math.equation(block: false, alt: "y")[$y$] direction. The end points of the beam satisfy the conditions, #math.equation(block: true, alt: "y open parenthesis 0 close parenthesis equals 0 , y double prime open parenthesis 0 close parenthesis equals 0 ,; y open parenthesis L close parenthesis equals 0 , y double prime open parenthesis L close parenthesis equals 0 .")[$y ( 0 ) = 0 , #h(2em) y^(″) ( 0 ) = 0 , \ y ( L ) = 0 , #h(2em) y^(″) ( L ) = 0 .$] See #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/5%3A_Eigenvalue_problems/5.2%3A_Application_of_Eigenfunction_Series")[Section 5.2], for further information about endpoint conditions applied to beams. #examplebox("Example 3")[][ Suppose that length of the beam is #math.equation(block: false, alt: "2")[$2$], and suppose that #math.equation(block: false, alt: "E I equals 1")[$E I = 1$] for simplicity. Further suppose that the force #math.equation(block: false, alt: "F equals 1")[$F = 1$] is applied at #math.equation(block: false, alt: "x equals 1")[$x = 1$]. That is, we have the equation #math.equation(block: true, alt: "the fraction d to the power 4 y over d x to the power 4 equals minus δ open parenthesis x minus 1 close parenthesis")[$frac(d^(4) y, d x^(4)) = − δ ( x − 1 )$] and the endpoint conditions are #math.equation(block: true, alt: "y open parenthesis 0 close parenthesis equals 0 , y double prime open parenthesis 0 close parenthesis equals 0 , y open parenthesis 2 close parenthesis equals 0 , y double prime open parenthesis 2 close parenthesis equals 0 .")[$y ( 0 ) = 0 , #h(1em) y^(″) ( 0 ) = 0 , #h(1em) y ( 2 ) = 0 , #h(1em) y^(″) ( 2 ) = 0 .$] We could integrate, but using the Laplace transform is even easier. We apply the transform in the #math.equation(block: false, alt: "x")[$x$] variable rather than the #math.equation(block: false, alt: "t")[$t$] variable. Let us again denote the transform of #math.equation(block: false, alt: "y open parenthesis x close parenthesis")[$y ( x )$] as #math.equation(block: false, alt: "Y open parenthesis s close parenthesis")[$Y ( s )$]. #math.equation(block: true, alt: "s to the power 4 Y open parenthesis s close parenthesis minus s cubed y open parenthesis 0 close parenthesis minus s squared y prime open parenthesis 0 close parenthesis minus s y double prime open parenthesis 0 close parenthesis minus y to the power ‴ open parenthesis 0 close parenthesis equals minus e to the power minus s .")[$s^(4) Y ( s ) − s^(3) y ( 0 ) − s^(2) y^(′) ( 0 ) − s y^(″) ( 0 ) − y^(‴) ( 0 ) = − e^(− s) .$] We notice that #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$] and #math.equation(block: false, alt: "y double prime open parenthesis 0 close parenthesis equals 0")[$y^(″) ( 0 ) = 0$]. Let us call #math.equation(block: false, alt: "C sub 1 equals y prime open parenthesis 0 close parenthesis")[$C_(1) = y^(′) ( 0 )$] and #math.equation(block: false, alt: "C squared equals y to the power ‴ open parenthesis 0 close parenthesis")[$C^(2) = y^(‴) ( 0 )$]. We solve for #math.equation(block: false, alt: "Y open parenthesis s close parenthesis")[$Y ( s )$], #math.equation(block: true, alt: "Y open parenthesis s close parenthesis equals the fraction minus e to the power minus s over s to the power 4 plus the fraction C sub 1 over s squared plus the fraction C sub 2 over s to the power 4 .")[$Y ( s ) = frac(− e^(− s), s^(4)) + frac(C_(1), s^(2)) + frac(C_(2), s^(4)) .$] We take the inverse Laplace transform utilizing the second shifting property Equation (6.2.14) to take the inverse of the first term. #math.equation(block: true, alt: "y open parenthesis x close parenthesis equals the fraction minus open parenthesis x minus 1 close parenthesis cubed over 6 u open parenthesis x minus 1 close parenthesis plus C sub 1 x plus the fraction C sub 2 over 6 x cubed .")[$y ( x ) = frac(− ( x − 1 )^(3), 6) u ( x − 1 ) + C_(1) x + frac(C_(2), 6) x^(3) .$] We still need to apply two of the endpoint conditions. As the conditions are at #math.equation(block: false, alt: "x equals 2")[$x = 2$] we can simply replace #math.equation(block: false, alt: "u open parenthesis x minus 1 close parenthesis equals 1")[$u ( x − 1 ) = 1$] when taking the derivatives. Therefore, #math.equation(block: true, alt: "0 equals y open parenthesis 2 close parenthesis equals the fraction minus open parenthesis 2 minus 1 close parenthesis cubed over 6 plus C sub 1 open parenthesis 2 close parenthesis plus the fraction C sub 2 over 6 2 cubed equals the fraction minus 1 over 6 plus 2 C sub 1 plus the fraction 4 over 3 C sub 2 ,")[$0 = y ( 2 ) = frac(− ( 2 − 1 )^(3), 6) + C_(1) ( 2 ) + frac(C_(2), 6) 2^(3) = frac(− 1, 6) + 2 C_(1) + frac(4, 3 C_(2)) ,$] and #math.equation(block: true, alt: "0 equals y double prime open parenthesis 2 close parenthesis equals the fraction minus 3 times 2 times open parenthesis 2 minus 1 close parenthesis over 6 plus the fraction C sub 2 over 6 3 times 2 times 2 equals minus 1 plus 2 C sub 2")[$0 = y^(″) ( 2 ) = frac(− 3 · 2 · ( 2 − 1 ), 6) + frac(C_(2), 6) 3 · 2 · 2 = − 1 + 2 C_(2)$] Hence #math.equation(block: false, alt: "C sub 2 equals the fraction 1 over 2")[$C_(2) = frac(1, 2)$] and solving for #math.equation(block: false, alt: "C sub 1")[$C_(1)$] using the first equation we obtain #math.equation(block: false, alt: "C sub 1 equals the fraction minus 1 over 4")[$C_(1) = frac(− 1, 4)$]. Our solution for the beam deflection is #math.equation(block: true, alt: "y open parenthesis x close parenthesis equals the fraction minus open parenthesis x minus 1 close parenthesis cubed over 6 u open parenthesis x minus 1 close parenthesis minus the fraction x over 4 plus the fraction x cubed over 12 .")[$y ( x ) = frac(− attach(( x − 1 ), t: 3), 6) u ( x − 1 ) − frac(x, 4) + frac(x^(3), 12) .$] ] === Footnotes \[1\] Named after the English physicist and mathematician Paul Adrien Maurice Dirac (1902–1984). \[2\] You should really think of the integral going over #math.equation(block: false, alt: "open parenthesis minus infinity , infinity close parenthesis")[$( − ∞ , ∞ )$] rather than over #math.equation(block: false, alt: "[ 0 , t ]")[$[ 0 , t ]$] and simply assume that #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] and #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$] are continuous and zero for negative . \[3\] Named for the Swiss mathematicians Jacob Bernoulli (1654–1705), Daniel Bernoulli —nephew of Jacob— (1700–1782), and Leonhard Paul Euler (1707–1783). \[4\] #math.equation(block: false, alt: "E")[$E$] is the elastic modulus and #math.equation(block: false, alt: "I")[$I$] is the second moment of area. Let us not worry about the details and simply think of these as some given constants.