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Module-5 Convolution in Time and Frequency Domain Objective

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Module-5 Convolution in Time and Frequency Domain Objective 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 ∞ 푗휆푡 −푗⍵푡 = [ x₁ λ 푒 dλ] x₂ t 푒 푑푡 −∞ 2휋 −∞ Interchanging the order of integration, we get 1 ∞ ∞ −푗⍵푡 푗휆푡 F[x₁(t) x₂(t)] = x₁ λ [ x₂ t 푒 푒 dt]dλ 2휋 −∞ −∞ 1 ∞ ∞ −푗(⍵−휆)푡 = x₁ λ [ x₂ t 푒 dt]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 −푎푡 −푏푡 = [푒 푢 푡 − 푒 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.
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