Discrete Time Convolution HW EEE 203 Wang

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Signals and Systems, I Discrete Time Convolution Signals and systems, I Homework: Discrete Time Convolution INTRODUCTION CONVOLUTION Convolution is a very important technique in signals and systems. While continuous-time convolution is important for theoretical analysis, you have to understand how to do discrete- time convolution in order to write a program to implement it on a computer or a DSP chip. This assignment will help you understand how to perform discrete time convolution operation with delta, window, unit step and exponential functions. For a discrete time, Linear and Time Invariant (LTI) system with impulse response h[n] as shown in Fig. 1, the output y[n] can be obtained through the convolution of input signal x[n] and impulse response h[n] as defined in Eqn. 1. TASK ONE: Convolution with a Delta Function Calculate the convolution of the following two signals 𝑦 [ 𝑛 ]= 𝑥 [ 𝑛 ] ∗ [ ℎ 𝑛 ] a. Write x[n] as a sum of delta functions similar to Eqn. 6. From the graph given: x [ n ] = ¿ Therefore: x [ n ] = σ [ n − 1 ] − σ [ n − 2 ] b. Compute convolution using the method shown in Eqn. 8. Plot your y[n] in a figure similar to Fig. 2. From the graph we are given that: x [ n ] = σ [ n − 1 ] − σ [ n − 2 ] h [ n ] = σ [ n ] + 2 σ [ n − 1 ] +− σ [ n − 2 ] From the convolution theory: y [ n ] = x [ n ] ∗ h [ n ] i h [ n ] ∗ σ [ n − n 0 ] = σ [ n − n 0 ] ii Applying equation i and ii we obtain:

Signals and Systems, I Discrete Time Convolution y [ n ] = h [ n ] ∗ σ [ n − 1 ] − h [ n ] ∗ σ [ n − 2 ] y [ n ] = σ [ n − 1 ] + σ [ n − 2 ] − 3 σ [ n − 3 ] + σ [ n − 4 ] This plot is as shown below: Figure 1: plot of the resultant signal from calculation. c. Go to http://jdsp.engineering.asu.edu/JDSP-HTML5/JDSP.html. Build the simulation diagram as shown in Fig. 1. Figure 2: Convolution as a sum of shifted and scaled input signals Open up Plot 6, 7 and 9. Choose “Plot Quantity” as “Real” at the top and “Plot” as “Discrete” at the bottom. Take a screen shot. Verify Plot 9 is a sum of Plot 6 and 7, as y[n] is a sum of scaled and shifted h[n]’s.

Signals and Systems, I Discrete Time Convolution Figure 3: Graph for plot 6 Figure 4: Graph for plot 7 Figure 5: Graph for plot 9

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Signals and Systems, I Discrete Time Convolution Compare your own plot y[n] with Plot 9. Are they the same? You can also see values of y[n] by clicking “Graph Values”. The results from calculation and simulation are the same. From the plotted graphs shown in figure 1 and figure 5. d. Verifying the results: Figure 6: Direct convolution Figure 7: Resultant waveform in plot 4

Signals and Systems, I Discrete Time Convolution TASK TWO Convolution with a Window Function Here a window function is informally defined as a finite number of impulses. Suppose the impulse response is a simple two-impulse window function: Calculate the convolution of the two signals 𝑦[𝑛] = 𝑥[𝑛] ∗ ℎ[𝑛] , where x[n] is defined in Fig. 7 and h[n] is defined in Eqn. 9. a. In Fig. 7, move the convolution machine from left to right and calculate y[n] by hand. Figure 8: Convolution machine with window function Solution: From the graph in figure above: x [ n ] = ¿ h [ n ] = 0.5 ( σ [ n ] + σ [ n − 1 ] )

Signals and Systems, I Discrete Time Convolution y [ n ] = x [ n ] ∗ h [ n ] = 0.5 ( x [ n ] + x [ n − 1 ] ) ……….i From equation I above therefore using the input x[n] we obtain the values of y[n] as: y [ 0 ] = 0.5 ∗ ( 0 + 1 ) = 0.5 ∗ 1 = 0.5 y [ 1 ] = 0.5 ∗ ( 1 + 3 ) = 4 ∗ 0.5 = 2 y [ 2 ] = 0.5 ∗ ( 3 + 2 ) = 0.5 ∗ 5 = 2.5 y [ 3 ] = 0.5 ∗ ( 2 + 1 ) = 0.5 ∗ 3 = 1.5 y [ 4 ] = 0.5 ∗ ( 0 + 1 ) = 0.5 ∗ 1 = 0.5 b. Go to http://jdsp.engineering.asu.edu/JDSP-HTML5/JDSP.html. Build the simulation diagram in Fig. 5. Define Sig. Gen. 1 and 2 as x[n] and h[n] using “User-Defined” signal. Open up Sig. Gen. 1, 2 and Plot 4, make sure all three plots are visible. Take a screenshot. (Take separate screenshots if they cannot fit on the same window.) Figure 9: Generator 1 signals Figure 10: Generator 2 signals

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Signals and Systems, I Discrete Time Convolution Figure 11: the output plot signals Figure 12: Graph values Is y[n] the same as you calculated in Part a? Yes, if you compare the calculated values in part a and the graph values in figure 12 above, they are the same. TASK THREE Convolution with a Shifted Window Function Now let’s look at a shifted version of the two-impulse window function:

Signals and Systems, I Discrete Time Convolution Figure 13: Convolution machine with time-shifted window function Calculate the convolution of the two signals 𝑦 [ 𝑛 ]= 𝑥 [ 𝑛 ] ∗ [ ℎ 𝑛 ], where x[n] is defined in Fig. 13 and h[n] is defined in Eqn. 11. a. In Fig. 8, move the convolution machine from left to right and calculate y[n] by hand. From the graph in figure 13 above: x [ n ] = ¿ h [ n ] = 0.5 ( u [ n − 2 ] − u [ n − 3 ] ) y [ n ] = 0.5 ( x [ n − 2 ] + x [ n − 3 ] ) Therefore, from the data given: N y = N x + N h − 1 = 2 + 4 − 1 = 5 y [ 2 ] = 0.5 ∗ ( 0 + 1 ) = 0.5 ∗ 1 = 0.5 y [ 3 ] = 0.5 ∗ ( 1 + 3 ) = 4 ∗ 0.5 = 2 y [ 4 ] = 0.5 ∗ ( 3 + 2 ) = 0.5 ∗ 5 = 2.5 y [ 5 ] = 0.5 ∗ ( 2 + 1 ) = 0.5 ∗ 3 = 1.5 y [ 6 ] = 0.5 ∗ ( 0 + 1 ) = 0.5 ∗ 1 = 0.5

Signals and Systems, I Discrete Time Convolution b. Go to http://jdsp.engineering.asu.edu/JDSP-HTML5/JDSP.html. Build the simulation diagram in Fig. 5. Define Sig. Gen. 1 and 2 as x[n] and h[n] using “User-Defined” signal. Open up Sig.Gen. 1, 2 and Plot 4, make sure all three plots are visible. Take a screenshot. (Take separate screenshots if they cannot fit on the same window.) Figure 14: Generator 1 signal output Figure 15: Signal generator 2 output Figure 16: Output signal data

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Signals and Systems, I Discrete Time Convolution Figure 17: Graph values Is y[n] the same as you calculated in Part a? yes they are the same. c. Suppose [ ℎ 𝑛 ]=12( 𝛿 [ 𝑛 ]+ 𝛿 [ 𝑛 +1]), derive y[n] in terms of x[n] similar to Eqn. 12 h [ n ] = 0.5 ( σ [ n ] + σ [ n + 1 ] ) y [ n ] = x [ n ] ∗ h [ n ] = ∑ k =− ∞ ∞ x [ k ] ∗ h [ n − k ] y [ n ]= 0.5 ∑ k =− ∞ ∞ x [ k ] ∗ ( σ [ n − k ] + σ [ n + 1 − k ] ) y [ n ]= 0.5 ∑ k =− ∞ ∞ x [ k ] ∗ ( σ [ n − k ] ) + 0.5 ∑ k =− ∞ ∞ x [ k ] ∗ ( σ [ n + 1 − k ] ) y [ n ] = 0.5 x [ n ] ∗ σ [ n ] + 0.5 x [ n ] ∗ σ [ n + 1 ] y [ n ]= 0.5 ( x [ n ] + x [ n + 1 ]) From calculation: y [ 0 ] = 0.5 ∗ ( 1 + 3 ) = 4 ∗ 0.5 = 2 y [ 1 ] = 0.5 ∗ ( 3 + 2 ) = 0.5 ∗ 5 = 2.5 y [ 2 ] = 0.5 ∗ ( 2 + 1 ) = 0.5 ∗ 3 = 1.5 y [ 3 ] = 0.5 ∗ ( 0 + 1 ) = 0.5 ∗ 1 = 0.5 Draw the convolution machine similar to Fig. 13.

Signals and Systems, I Discrete Time Convolution Figure 18: Resultant output signal Is the system causal? Why? No, the system dependents only on the present values and the next values. TASK FOUR Convolution with a Unit Step Function Now what if the impulse response is a unit step function as shown in Eqn. 13? First convince yourself the unit step function is simply a sum of infinite number of shifted impulses as shown in Eqn. 13. By applying the sifting property of Eqn. 2 and 3, we have: Eqn. 14 shows y[n] is the sum of all current and previous input x[n]’s. Calculate the convolution of the two signals 𝑦 [ 𝑛 ] = 𝑥 [ 𝑛 ] ∗ [ ℎ 𝑛 ], where x[n] is defined in Fig. 19 and h[n] is defined in Eqn. 13.

Signals and Systems, I Discrete Time Convolution Figure 19: Convolution machine with unit step function. a. In Fig. 19, move the convolution machine from left to right and calculate y[n] by hand. x [ n ] = ¿ h [ n ] = ∑ i = 0 ∞ σ [ n − i ] y [ n ] = ∑ i = 0 ∞ x [ n − i ] Therefore: y [ 0 ] = 1 y [ 1 ] = 3 + 1 = 4 y [ 2 ] = 2 + 3 + 1 = 6 y [ 3 ] = 1 + 2 + 3 + 1 = 7 y [ n≥ 3 ] = 7 b. Go to http://jdsp.engineering.asu.edu/JDSP-HTML5/JDSP.html. Build the simulation diagram in Fig. 5. Define Sig. Gen. 1 as x[n] using “User-Defined” signal. Define Sig. Gen. 2 as h[n] using “Rectangular” signal. Set “Gain” to 1, “Pulse width” to 20 (from

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Signals and Systems, I Discrete Time Convolution 0 to 19), “Time Shift” to 0. Open up Sig. Gen. 1, 2 and Plot 4, make sure all three plots are visible. Take a screenshot. (Take separate screenshots if they cannot fit on the same window.) Figure 20: Signal generator 1 output signal Figure 21: Signal generator 2 output signals Figure 22: Output signal of the plot Is y[n] the same as you calculated in Part a? Why does y[n] start to drop at n = 20? Yes, they are the same. This is because the convoluting signal has been only up to n=20. TASK FIVE Convolution with an Exponential Function

Signals and Systems, I Discrete Time Convolution When the impulse response is an exponential function such as: it can be decomposed into a sum of infinite number of impulses like the unit step function. 𝛿 [ 𝑛 − 𝑖 ] represents the location of the impulse and (12) 𝑖 represents its value. By applying the sifting property of Eqn. 2 and 3, we have: Eqn. 16 shows y[n] is the sum of all current and previous input x[n]’s scaled by the (12) 𝑖 factor. Figure 23: Convolution machine with exponential function Calculate the convolution of the two signals 𝑦 [ 𝑛 ] = 𝑥 [ 𝑛 ] ∗ [ ℎ 𝑛 ], where x[n] is defined in Fig. 20 and h[n] is defined in Eqn. 15. a. In Fig. 10, move the convolution machine from left to right and calculate y[n] by hand for n =0 to 4.

Signals and Systems, I Discrete Time Convolution x [ n ] = ¿ h [ n ] = ∑ i = 0 ∞ ( 1 2 ) i σ [ n − i ] y [ n ] = ∑ i = 0 ∞ ( 1 2 ) i x [ n − i ] Therefore: y [ 0 ] = x [ 0 ] = 1 y [ 1 ] = x [ 0 ] 2 + x [ 1 ] = 3.5 y [ 2 ] = x [ 0 ] 4 + x [ 1 ] 2 + x [ 2 ] = 3.75 y [ 3 ] = x [ 0 ] 8 + x [ 1 ] 4 + x [ 2 ] 2 + x [ 3 ] = 2.875 y [ 4 ] = x [ 0 ] 16 + x [ 1 ] 8 + x [ 2 ] 4 + x [ 3 ] 4 + x [ 4 ] = 1.4375 y [ 4 ] = x [ 0 ] 32 + x [ 1 ] 16 + x [ 2 ] 8 + x [ 3 ] 4 + x [ 4 ] 2 x [ 5 ] = 0.71875 b. Go to http://jdsp.engineering.asu.edu/JDSP-HTML5/JDSP.html. Build the simulation diagram in Fig. 5. Define Sig. Gen. 1 as x[n] using “User-Defined” signal. Define Sig. Gen. 2 as h[n] using “Exponential” signal. Set “Exp Base” to 0.5, “Gain” to 1, “Pulse width” to 20 (from 0 to 19), “Time Shift” to 0. Open up Sig. Gen. 1, 2 and Plot 4, make sure all three plots are visible. Take a screenshot. (Take separate screenshots if they cannot fit on the same window.) Figure 24: Signal generator 1 output signal

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Signals and Systems, I Discrete Time Convolution Figure 25: Signal generator 2 output signal Figure 26: Plot 4 output signal Is y[n] the same as you calculated in Part a? Open up “Graph Values” in Plot 4 will help you compare values. Yes, they are the same as shown in figure 27. Figure 27: Graph values