Module-5 Convolution in time and frequency domain
Objective : To understand the importance of Convolution operation in LTI systems.
Introduction: In this lecture we prove the most important theorem regarding the Fourier Transform –the Convolution Theorem. It is the theorem that links the Fourier Transform to LSI systems, and opens up a wide range of application to Fourier Transform.
Description: Convolution Theorems: Convolution of signals may be done either in time domain or frequency domain. So there are following two theorems of convolution associated with Fourier transforms: 1.Time convolution theorem 2.Frequency convolution theorem Time convolution theorem The time convolution theorem states that convolution in time domain is equivalent to multiplication of their spectra in frequency domain. Mathematically, if x1(t)↔X1(⍵) and x2(t)↔X2(⍵) then x1(t) * x2(t)↔ X1(⍵)X2(⍵) ∞ Proof: F[x (t) * x (t)] [x₁(t) ∗ x₂(t)] 푒−푗푤푡 푑푡 1 2 = −∞ ∞ we have x (t) * x (t) = x₁(τ) x₂(t − τ) 푑흉 1 2 −∞ ∞ ∞ F[x (t) * x (t)] = { [x₁(τ) x₂(t − τ)dτ] 푒−푗푤푡 푑푡} 1 2 −∞ −∞ Interchanging the order of integration, we have ∞ ∞ F[x (t) * x (t) ]= x₁(τ) [ x₂(t − τ) 푒−푗푤푡 푑푡]푑휏 1 2 −∞ −∞ Letting t-휏 = p, in the second integration, we have t=p+휏 and dt = dp ∞ ∞ F[x (t) * x (t)] = x₁(τ) [ x₂(p) 푒−푗푤 (푝+휏)푑푝]푑휏 1 2 −∞ −∞ ∞ ∞ = x₁(τ) [ x₂(p) 푒−푗푤푝 푑푝]푒−푗푤휏 푑휏 −∞ −∞ ∞ ∞ = x₁ τ 푋₂(휔)푒−푗푤휏 푑휏 = x₁ τ 푒−푗푤휏 푑휏푋₂(휔) −∞ −∞ = X₁(⍵)X₂(⍵) x₁(t) * x₂(t)↔ X₁(⍵)X₂(⍵) This is time convolution theorem. Frequency convolution theorem The frequency convolution theorem states that the multiplication of two functions in time domain is equivalent to convolution of their spectra in frequency domain. Mathematically, if x₁(t)↔X₁(⍵) and x₂(t)↔X₂(⍵) 1 then x₁(t) x₂(t)↔ [ X₁(⍵) * X₂(⍵)] 2휋 ∞ Proof: F[x₁(t) x₂(t)] = [x₁(t) x₂(t)] 푒−푗⍵푡푑푡 −∞ ∞ 1 ∞ = [ x₁ λ 푒푗휆푡 dλ] x₂ t 푒−푗⍵푡 푑푡 −∞ 2휋 −∞ Interchanging the order of integration, we get 1 ∞ ∞ F[x₁(t) x₂(t)] = x₁ λ [ x₂ t 푒−푗⍵푡푒푗휆푡 dt]dλ 2휋 −∞ −∞ 1 ∞ ∞ = x₁ λ [ x₂ t 푒−푗(⍵−휆)푡dt]dλ 2휋 −∞ −∞ 1 ∞ = x₁ λ 푥₂(⍵ − 휆) 푑휆 2휋 −∞ 1 = [ X₁(⍵) * X₂(⍵)] 2휋 1 x₁(t) x₂(t)↔ [ X₁(⍵) * X₂(⍵)] 2휋 2π x₁(t) x₂(t) ↔ X₁(⍵) * X₂(⍵) This is frequency convolution theorem in radian frequency in terms of frequency, we get F[x₁(t) x₂(t)]= X₁(f) * X₂(f) Illustration
Example: Find the convolution of the signals x₁(t) = 푒−푎푡 푢(푡) ; x₂(t)=푒−푏푡 푢(푡) using Fourier transform. Solution: Given x₁(t) = 푒−푎푡 푢(푡) 1 x₁(ω)= 푎+푗휔 −푏푡 x2(t)=푒 푢(푡) 1 x₂(ω)= b+jω we know that F[x₁(t) * x₂(t)] = X₁(⍵)X₂(⍵) x₁(t) * x₂(t) = 퐹−1 [X₁(⍵)X₂(⍵)] 1 1 1 1 x₁(t) * x₂(t)= 퐹−1[ ] = 퐹−1[ ( − )] 푎+푗휔 푏+푗휔 푏−푎 푎+푗휔 푏+푗휔 1 1 1 = [퐹−1 − 퐹−1 ] 푏−푎 푎+푗휔 푏+푗휔 1 = [푒−푎푡 푢 푡 − 푒−푏푡 푢 푡 ] 푏−푎 Example: Find the convolution of the signals x₁(t) = 2푒−2푡푢(푡) ; x₂(t)=푢(푡) using Fourier transform Solution: Given x₁(t) = 2푒−2푡푢(푡) 2 x₁(ω)= 2+jω x₂(t)=u(t) 1 x₂(ω)=휋δ ω + jω 2 1 2 2( πδ ω x₁(ω)x₂(ω) = ( πδ ω + ) = + 2+jω jω jω(2+jω) 2+jω since x₁(t) * x₂(t) = 퐹−1 [X₁(⍵)X₂(⍵)] , we have 2 2휋훿 휔 1 1 2 휋훿 휔 x₁(t) * x₂(t) = 퐹−1 [ + ] = 퐹−1 [ − + ] 푗휔 2+푗휔 2+푗휔 푗휔 푗휔 +2 2+푗휔 2 휋훿 휔 since δ(ω)=1 for ω=0 and δ(ω)=0 for ω not equals to zero , we have = 2+푗휔 휋훿 휔 1 1 x₁(t) * x₂(t) = 퐹−1 [ + 휋훿 휔 − ] 푗휔 푗휔 +2 1 1 =퐹−1 + 휋훿 휔 − 퐹−1 ( ) 푗휔 푗휔 +2 = u(t) - 푒−2푡푢(푡) = (1-푒−2푡) u(t)
Graphical representation of convolution: The convolution of two signals can be performed using graphical method. The procedure is: 1. For the given signals x(t) and h(t), replace the independent variable t by a dummy variable 휏 and plot the graph for x(휏) and h(휏). 2. keep the function x(휏) fixed. visualize the function h(휏) as a rigid wire frame and rotate (or invert) this frame about the vertical axis (휏 = 0) to obtain h(-휏). 3. shift the frame along the 휏-axis by t sec. the shifted frame now represents h(t-휏). 4. plot the graph for x(휏) and h(t-휏) on the same axis beginning with very large negative time shift t. 5. for a particular value of t=a , integration of the product x(휏)h(t-휏) represents the area under the product curve (common area). this common area represents the convolution of x(t) and h(t) for a shift of t=a. ∞ i.e., 푥 휏 푕(푎 − 휏)푑휏 =[x(t)* h(t)] −∞ 6. Increase the time shift t and take the new interval whenever the function either x(휏) or h(t-휏) changes. the value of t at which the change occurs defines the end of the current interval and the beginning of a new interval. calculate y(t) using step5. 7. the value of convolution obtained at different values of t ( both positive and negative values) may be plotted on a graph to get the combined convolution. Illustration Example : Find the convolution of the signals by graphical method x(t) = 푒−3푡푢(푡) ; h(t)=푢(푡 + 3) Solution: Given x(t) = 푒−3푡푢(푡) ; h(t)=푢(푡 + 3) ∞ The output y(t)= x(t) * h(t) = 푥 휏 푕(푡 − 휏)푑휏 −∞ The two functions x(휏) and h(휏) will be x(휏) = 푒−3휏 푢(휏) = 푒−3휏 for 휏 ≥0 h(휏)=푢(휏 + 3)=1 for 휏≥ -3 h(-휏)=푢(−휏 + 3) h(-휏) can be obtained by folding h(휏) about 휏=0. Figure shows the plots of x(휏) and h(휏).
Figure shows the plots of x(τ) and h(t-τ) together on the same time axis. Here the signal h(t-τ) is sketched for t<-3. x(τ) and h(t-τ) do not overlap. Therefore, the product x(τ) h(t-τ) is equal to zero. y(t)=0 (for t<-3)
Plots of (a) x(휏), and (b)h(t-휏)when there is no overlap Now, increase the time shift t until the signal h(t-τ) intersects x(τ). Figure below shows the situation for t>-3. Here x(τ) and h(t-τ) overlapped.[ This overlapping continuous for all values for t>-3 up to t=∞ because x(τ) exists for all values of τ>0]. But x(τ)=0 to τ=t+3.
Plot of x(휏), and h(t-휏) with overlap t+3 y(t)= x τ h(t − τ)dτ 0 푡+3 = 푒−3휏 푑휏 =[ 푒−3휏/-3]₀t+3 = [푒−3 푡+3 − 1]/−3 = 1 − 푒−3 푡+3 /3 0 y(t)= 0 for t<-3 = 1 − 푒−3 푡+3 /3 for t>-3
Convolution properties of Fourier transform: With two functions h(t) and g(t), and their corresponding Fourier transforms H(f) and G(f), we can form two special combinations – The convolution, denoted f = g * h, defined by ∞ f(t) = g ∗h ≡ 푔 휏 푕(푡 − 휏) 푑휏 −∞ Convolution: g*h is a function of time, and g*h = h*g – The convolution is one member of a transform pair g*h ↔ G(f) H(f) The Fourier transform of the convolution is the product of the two Fourier transforms! – This is the Convolution Theorem. Worked out Problems:
Example 1: the input and impulse response to the system are given by
x(t)=u(t+2)
h(t)=u(t-3)
Determine the output of the system graphically.
Solution: Given x(t)=u(t+2) and h(t)=u(t-3). The output y(t) is convolution of x(t) and h(t)
∞ y(t) = x(t)* h(t)= 푥 휏 푕(푡 − 휏)푑휏 −∞ The two functions x(휏) and h(휏) will be x(휏)=u(휏+2)=1 (for 휏≥ -2)
h(휏)=u(휏-3)=1 (for 휏≥ 3)
The functions x(휏) , h(휏) and h(-휏) are plotted as shown in fig
h(-τ
Plots of (a) x(τ), (b) h(τ) and (c) h(-τ)
Below figure shows the plots of the functions x(휏) and h(t-휏) together on the same axis. h(t-휏) is sketched for t-3 < -2 , i.e., for t<1 .
Plots of x(τ) and h(t-τ) for t < 1
For t<1, x(휏)and h(t-휏) do not overlap because
x(휏)=0 (for 휏 < -2)
h(t-휏)=0 (for t-3 <-2)
y(t)=0
Below figure shows the plots of x(휏) and h(t-휏) when t-3 < -2 or t>1. Now, there is an overlap between signals x(휏) and h(t-휏) in the interval -2 <휏< t-3.
Plots of x(τ) and h(t-τ) when there is ovelap.
From the above figure , we have 3 separate regions as follows:
For -∞ <휏<-2 , x(휏)h(t-휏)=0 ,since there is no overlap.
For -2<휏< t-3 , x(휏)h(t-휏)≠0 ,since there is an overlap.
For t-3 ≤휏 ≤ ∞ , x(휏)h(t-휏)=0 ,since there is no overlap.
Based on the above, we can write the convolution integral as :
∞ y(t)= x(τ) h(t − τ)d휏 −∞ −2 푡−3 ∞ = 0 ⨯ 1 푑휏 + 1 ⨯ 1 푑휏 + 1 ⨯ 0 푑휏 −∞ −2 푡−3
푡−3 = 푑휏 = [휏] -₂t-3 −2
= t-3+2 =t-1
y(t)=0 (for t <1)
= t-1 (for t >1)
This function ca n also be written as:
y(t) = (t-1) u(t-1) this is unit ramp delayed by 1. Its plot is shows in figure below.
Plots of y(t) = u(t+2)*u(t-3)
Example2 :The impulse response of the circuit is given as h(t) = 푒−2푡푢 푡 . This circuit is excited by an input of x(t) = 푒−4푡 푢 푡 − 푢 푡 − 2 . Determine the output of the circuit. Solution: Here the impulse response and input are:
h(t)= 푒−2푡푢 푡 = 푒−2푡 (for t ≥ 0)
x(t) = 푒−4푡 푢 푡 − 푢 푡 − 2 =푒−4푡 (for 0 ∞ y(t)= x(t)* h(t) 푥 휏 푕(푡 − 휏)푑휏 = −∞ Writing x(t) and h(t)in terms of 휏, we have x(휏) = 푒−4휏 for 0 ≤ 휏 ≤ 2 h(휏)=푒−2휏 (for 휏 ≥ 0 Below figure shows the plots of x(휏) , h(휏) and h(-휏) w.r.t 휏. Plots of (a)x(휏) ,(b) h(휏) and (c)h(-휏) For the convolution of x(t) and h(t), we require h(t-휏). h(t-휏) = 푒−2 푡−휏 푢 푡 − 휏 = 푒−2 푡−휏 (for t-휏 > 0 or 휏 The plots of x(휏) and h(t-휏) drawn on the same time axis are shown in the below figure for t<0. The plots do not overlap. y(t) =0 (for t<0) Plots of x τ , and h t-τ for t<0 For 0≤ t ≤2 Below figure shows the plots of x(휏) and h(t-휏) For 0≤ t ≤2 drawn on the same time axis. Observe that there is an overlap between x(휏) and h(t-휏) as shown by the shaded area only for 0 to t. Plots of x(τ) and h(t-τ) when there is an overlap For 0≤ t ≤2 we can write the convolution as : 0 푡 2 y(t)= 0 ⨯ 푕(푡 − 휏)푑휏 + 푥 휏 푕 푡 − 휏 푑휏 + 푥 휏 ⨯ (0)푑휏 −∞ 0 푡 푡 = 푥 휏 푕 푡 − 휏 푑휏 0 푡 = (푒−4휏 ) (푒−2 푡−휏 ) 푑휏 0 푡 = 푒−2푡 푒−2휏 푑휏 0 푒 −2휏 푒 −2푡−1 1 = 푒−2푡 ₒᵗ = 푒−2푡 = 푒−2푡 1 − 푒−2푡 −2 −2 2 1 y(t)= 푒−2푡 1 − 푒−2푡 (For 0≤ t ≤2) 2 For t > 2 Now, consider the case t > 2. For t > 2, the plots of x(휏) and h(t-휏) drawn on the same time axis are shown in below figure. In this figure, observe that x(휏) and h(t-휏) overlap only for 0 ≤ t ≤ 2 as shown by the shaded area. Plots of x(τ), and h(t-τ) for t > 2 Hence we can write the convolution equation as : 0 2 푡 y(t)= 0 ⨯ 푕(푡 − 휏)푑휏 + 푥 휏 푕 푡 − 휏 푑휏 + 0 ⨯ 푕(푡 − 휏)푑휏 −∞ 0 2 2 2 2 = 푥 휏 푕 푡 − 휏 푑휏 = (푒−4휏 ) (푒−2 푡−휏 ) 푑휏 = 푒−2푡 푒−2휏 푑휏 0 0 0 푒 −2휏 푒 −4−1 = 푒−2푡 ₒ²= 푒−2푡 −2 −2 1 = 푒−2푡 1 − 푒−4 for t > 2 2 Thus, we obtained the convolution as follows : 0 푓표푟 푡 < 0 1 푒−2푡 1 − 푒−2푡 푓표푟 0 < 푡 < 2 y(t)= 2 1 푒−2푡 1 − 푒−4 푓표푟 푡 > 2 2 This function is plotted in the below figure : 1 At t=2, the value of y(t)= 푒−4 1 − 푒−4 =0.009 2 In the above figure, observe that y(t) increases from t=0 to t=2. It has the maximum value at t=2. Then y(t) decays exponentially. −푎푡 −푏푡 Example 3 : Find the convolution of the signals x1(t) = 푒 u(t) ; x2(t)= 푒 u(t) using Fourier transform. −푎푡 Solution: Given x1(t) = 푒 u(t) 1 X1(ω)= 푎+푗휔 −푏푡 x2(t) = 푒 u(t) 1 X2(ω)= 푏+푗휔 F[x1(t) * x2(t)] = X1(ω)X2(ω) −1 x1(t) * x2(t) = 퐹 [X1(ω)X2(ω)] −1 1 −1 1 1 1 x1(t) * x2(t) = 퐹 [ ] = 퐹 − (푎+푗휔)(푏+푗휔) 푏−푎 푎+푗휔 푏+푗휔 1 1 1 = 퐹−1 − 퐹−1 푏−푎 푎+푗휔 푏+푗휔 1 = 푒−푎푡 u t − 푒−푏푡 u(t) 푏−푎 −2푡 Example 4 : Find the convolution of the signals x1(t) = 2푒 u(t) ; x2(t)= u(t) using Fourier transform. −2푡 Solution: Given x1(t) = 2푒 u(t) 2 X1(ω)= 2+푗휔 x2(t) = u(t) 1 X2(ω)= 휋훿 휔 + 푗휔 2 1 2 2휋훿 휔 X1(ω)X2(ω) = 휋훿 휔 + = + 2+푗휔 푗휔 푗휔(2+푗휔) 2+푗휔 −1 Since x1(t) * x2(t) = 퐹 [X1(ω)X2(ω)] , we have −1 2 2휋훿 휔 −1 1 1 2휋훿 휔 x1(t) * x2(t) = 퐹 + = 퐹 − + 푗휔(2+푗휔) 2+푗휔 푗휔 (2+푗휔) 2+푗휔 2휋훿 휔 Since 훿 휔 = 1 for ω=0 and 훿 휔 = 0 for ω ≠ 0 , we have = 휋훿 휔 . 2+푗휔 −1 1 1 −1 1 −1 1 x1(t) * x2(t) = 퐹 + 휋훿 휔 − = 퐹 + 휋훿 휔 − 퐹 푗휔 (2+푗휔) 푗휔 (2+푗휔) = u(t) - 푒−2푡u(t) = (1- 푒−2푡)u(t) −푡 −푡 Example 5 :Find the convolution of the signals x1(t) = 푒 u(t) ; x2(t)= 푒 u(t) using Fourier transform. −푡 Solution : Given x1(t) = 푒 u(t) 1 X1(ω)= 1+푗휔 −푡 x1(t) = 푒 u(t) 1 X1(ω)= 1+푗휔 1 1 1 X1(ω)X2(ω) = = 1+푗휔 1+푗휔 (1+푗휔)2 −1 Since x1(t) * x2(t) = 퐹 [X1(ω)X2(ω)] , we have −1 1 −푡 x1(t) * x2(t) = 퐹 = 푡푒 u(t) (1+푗휔)2 −2푡 Example 6: using Fourier transform, find the convolution of the signals x1(t) = 푒 u(t) ; x2(t)= 푒−3푡 u(t) Solution:The convolution property of Fourier transform says that x1(t) * x2(t) ↔ X1 (ω) X2 (ω) −1 x1(t) * x2 (t) = 퐹 [ X1 (ω) X2 (ω)] −2푡 x1 (t)= 푒 u(t) 1 X1(ω) = 2+jω −3푡 x2(t)= 푒 u(t) 1 X2(ω) = 3+jω 1 x (t) * x (t) = 퐹−1 [ X (ω) X (ω)] = 퐹−1 1 2 1 2 (jω+2)(3+jω) 1 A B 1 1 = = + = - (jω+2)(3+jω) 2+jω 3+jω 2+jω 3+jω 1 1 x (t) * x (t) = 퐹−1 − = 푒−2푡u(t) - 푒−3푡 u(t) 1 2 2+jω 3+jω −푡 Example 7:using Fourier transform, find the convolution of the signals x1(t) = t푒 u(t) ; x2(t)= t푒−2푡 u(t) −푡 Solution: x1(t) = t푒 u(t) 1 X (ω) = 1 1+jω 2 −2푡 x2(t)= t푒 u(t) 1 X (ω) = 2 2+jω 2 x1(t) * x2(t) ↔ X1 (ω) X2 (ω) 1 x (t) * x (t) = 퐹−1 [ X (ω) X (ω)] = 퐹−1 1 2 1 2 1+jω 2 2+jω 2 Taking partial fractions, we have 1 A B c 퐷 X(ω)= = + + + 1+jω 2 2+jω 2 1+jω 1+jω 2 2+jω 2+jω 2 −2 1 2 1 = + + + 1+jω 1+jω 2 2+jω 2+jω 2 푥(푡) = −2푒−푡 푢(푡) + 푡 푒−푡 푢(푡) + 2. 푒−2푡 푢(푡) + 푡 푒−2푡 푢(푡) Example 8:Find the response of the system whose impulse response is h(t)=δ(t) for the input x(t) = e-2tu(t) Solution: Given impulse response h(t)=δ(t) Input x(t) = e-2tu(t) Output y(t)=x(t) * h(t) There y(t) = e-2tu(t) * δ(t) =e-2tu(t) Example 9:Find the net impulse response of the two systems connected in cascade with impulse responses h1(t)= δ(t) and h2(t)=u(t). Solution: Net impulse response is obtained by convolving h1(t) and h2(t) h(t)= h1(t)*h2(t)=δ(t)*u(t)= u(t) Example 10: Find the transfer function of the system mentioned in Example 9 Solution: 1 H(ω)=휋훿 휔 + 푗휔 Assignment Questions: −푡 −3푡 1. Find the convolution of the signal x1(t) = 푒 u(t) ; x2(t)= 푒 u(t) 2. Find the convolution of the signal x1(t) = u(t) ; x2(t)= u(t) 3. Find the convolution of the signal x1(t) = sin 푡u(t) ; x2(t)= u(t) −푡 4. Find the convolution of the signal x1(t) = 푒 u(t) ; x2(t)= u(t) −2푡 −4푡 5. Find the convolution of the signals x1(t) = 2푒 u(t) ; x2(t) = 푒 u(t) using Fourier transform. −푡 −푡 6. Find the convolution of the signals x1(t) = 푒 u(t) ; x2(t) = 푒 u(t) using Fourier transform. −푎푡 7. Find the convolution of the signal by graphical method. x1(t) = 푒 u(t) ; x2(t)= u(t) 8. Find the convolution of the signal by graphical method. x1(t) = u(t + 1) ; x2(t)= u(t-2) −푡 9. Find the convolution of the signal by graphical method. x1(t) = 푒 u(t) ; −3푡 x2(t)= 푒 [u(t) – u(t-2)] −3푡 10. Find the convolution of the signal by graphical method. x1(t) = 푒 u(t) ; x2(t)= u(t-3) – u(t-5) Simulation: %FINDING THE CONVOLUTION BETWEEN 푒−2푡 and 푢(푡) between 0 < t < 1 clc; clear all; close all; t=linspace(0,1,1000); x=exp(-2*t); h=heaviside(t); l=length(x); %Finding lengths m=length(h); subplot(3,1,1); %plotting the input sequences plot(0:l-1,x); xlabel('t---->'); ylabel('x(t)--->'); subplot(3,1,2); plot(0:m-1,h); xlabel('t---->'); ylabel('h(t)--->'); y=conv(x,h); disp('The convolution result is '); y %Displaying the resultant sequence in command window subplot(3,1,3); %Plotting the resultant sequence plot(0:length(y)-1,y); xlabel('t---->'); ylabel('y(t)=x(t)*h(t)--->'); title('Convoluted result'); References: [1] Alan V.Oppenheim, Alan S.Willsky and S.Hamind Nawab, “Signals & Systems”, Second edition, Pearson Education, 8th Indian Reprint, 2005. [2] M.J.Roberts, “Signals and Systems, Analysis using Transform methods and MATLAB”, Second edition,McGraw-Hill Education,2011 [3] John R Buck, Michael M Daniel and Andrew C.Singer, “Computer explorations in Signals and Systems using MATLAB”,Prentice Hall Signal Processing Series [4] P Ramakrishna rao, “Signals and Systems”, Tata McGraw-Hill, 2008 [5] Tarun Kumar Rawat, “Signals and Systems”, Oxford University Press,2011 [6]A.Anand Kumar, “Signals and Systems” , PHI Learning Private Limited ,2011