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Module 2 Problems-1

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Module 2 Problems
Evaluate the discrete – time convolution sum of the following
1. 𝑦[𝑛] = 𝑒[𝑛] ∗ 𝑒[𝑛 − 3]
2. π‘₯[𝑛] = 𝛼 𝑛 𝑒[𝑛]; β„Ž[𝑛] = 𝑒[𝑛]
1 𝑛
3. 𝑦[𝑛] = (2) 𝑒[𝑛 − 2] ∗ 𝑒[𝑛]
4. 𝑦[𝑛] = (𝛽)𝑛 𝑒[𝑛] ∗ 𝑒[𝑛 − 3]
5. 𝑦[𝑛] = (𝛽)𝑛 𝑒[𝑛] ∗ 𝛼 𝑛 𝑒[𝑛]
Consider a input signal x[n] and impulse response h[n] given as
1; 0≤𝑛 ≤ 4
π‘₯[𝑛] = {
0; π‘‚π‘‘β„Žπ‘’π‘Ÿπ‘€π‘–π‘ π‘’
𝛼𝑛 ; 0 ≤ 𝑛 ≤ 6
β„Ž[𝑛] = {
0; π‘‚π‘‘β„Žπ‘’π‘Ÿπ‘€π‘–π‘ π‘’
𝛼>1
Compute the output signal y[n]. Also plot y[n] for 𝛼 = 2
Evaluate the following
1. 𝑦(𝑑) = 𝑒(𝑑 + 1) ∗ 𝑒(𝑑 − 2)
2. 𝑦(𝑑) = 𝑒 −2𝑑 𝑒(𝑑) ∗ 𝑒(𝑑 + 2)
3. 𝑦(𝑑) = {𝑒(𝑑 + 2) − 𝑒(𝑑 − 1)} ∗ 𝑒(−𝑑 + 2)
A discrete – time LTI system is characterized by impulse response. Determine whether the
impulse response is (a) Stable and (b) causal.
1 𝑛
1. β„Ž[𝑛] = (2) 𝑒[𝑛]
2. β„Ž[𝑛] = (0.99)𝑛 𝑒[𝑛 + 3]
3. β„Ž[𝑛] = (4)−𝑛 𝑒[2 − 𝑛]
A continuous – time LTI system is characterized by impulse response. Determine whether
the impulse response is (a) Stable and (b) causal.
1. β„Ž(𝑑) = 𝑒 −3𝑑 𝑒(𝑑 − 1)
2. β„Ž(𝑑) = 𝑒 −𝑑 𝑒(𝑑 + 100)
3. β„Ž(𝑑) = 𝑒 𝑑 𝑒(−𝑑 − 1)
4. β„Ž(𝑑) = 𝑒 −4|𝑑|
Find the step response for the LTI system represented by the impulse response
1 𝑛
1. β„Ž[𝑛] = (2) 𝑒[𝑛]
2. β„Ž(𝑑) = 𝑑 𝑒(𝑑)
3. β„Ž(𝑑) = 𝑒 −|𝑑|
4. β„Ž(𝑑) = 𝛿(𝑑) − 𝛿(𝑑 − 1)
Determine convolution of the two given sequences
1. π‘₯[𝑛] = {1, ⏟
2 , 3, 4} π‘Žπ‘›π‘‘ β„Ž[𝑛] = {1
⏟ , 1, 3, 2}
2. π‘₯[𝑛] = {1
⏟ , 2, 4} π‘Žπ‘›π‘‘ β„Ž[𝑛] = {1
⏟ , 1, 1, 1, 1, 1}
Sketch the direct form I and direct form II implementations for the following
1
1. 𝑦[𝑛] + 2 𝑦[𝑛 − 1] − 𝑦[𝑛 − 3] = 3π‘₯[𝑛 − 1] + 2π‘₯[𝑛 − 2]
1
1
1
2. 𝑦[𝑛] − 4 𝑦[𝑛 − 1] + 8 𝑦[𝑛 − 2] = π‘₯[𝑛] + 2 π‘₯[𝑛 − 2]
3.
𝑑 3 𝑦(𝑑)
𝑑𝑑
+2
𝑑 2 𝑦(𝑑)
𝑑𝑑
+ 3𝑦(𝑑) = π‘₯(𝑑) + 3
𝑑π‘₯(𝑑)
𝑑𝑑
Solve the following homogeneous differential and difference equations with the specified
initial conditions.
1.
2.
3.
𝑑 2 𝑦(𝑑)
𝑑𝑑
𝑑 2 𝑦(𝑑)
𝑑𝑑
𝑑 3 𝑦(𝑑)
𝑑𝑑
+3
𝑑𝑦(𝑑)
+2
𝑑𝑦(𝑑)
+
𝑑𝑑
𝑑𝑑
𝑑 2 𝑦(𝑑)
𝑑𝑑
+ 2𝑦(𝑑) = 0
+ 𝑦(𝑑) = 0
−
𝑑𝑦(𝑑)
𝑑𝑑
;
;
𝑦(0) = 0 ;
𝑦(0) = 1 ;
𝑦 ′ (0) = 2
𝑦 ′ (0) = 1
− 𝑦(𝑑) = 0 ; 𝑦(0) = 1 ; 𝑦 ′ (0) = 1; 𝑦 ′′ (0) = −2
4. 𝑦[𝑛] − 3𝑦[𝑛 − 1] − 4𝑦[𝑛 − 2] = 0
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