Module 2 Problems
Evaluate the discrete – time convolution sum of the following
1. π¦[π] = π’[π] ∗ π’[π − 3]
2. π₯[π] = πΌ π π’[π]; β[π] = π’[π]
1 π
3. π¦[π] = (2) π’[π − 2] ∗ π’[π]
4. π¦[π] = (π½)π π’[π] ∗ π’[π − 3]
5. π¦[π] = (π½)π π’[π] ∗ πΌ π π’[π]
Consider a input signal x[n] and impulse response h[n] given as
1; 0≤π ≤ 4
π₯[π] = {
0; ππ‘βπππ€ππ π
πΌπ ; 0 ≤ π ≤ 6
β[π] = {
0; ππ‘βπππ€ππ π
πΌ>1
Compute the output signal y[n]. Also plot y[n] for πΌ = 2
Evaluate the following
1. π¦(π‘) = π’(π‘ + 1) ∗ π’(π‘ − 2)
2. π¦(π‘) = π −2π‘ π’(π‘) ∗ π’(π‘ + 2)
3. π¦(π‘) = {π’(π‘ + 2) − π’(π‘ − 1)} ∗ π’(−π‘ + 2)
A discrete – time LTI system is characterized by impulse response. Determine whether the
impulse response is (a) Stable and (b) causal.
1 π
1. β[π] = (2) π’[π]
2. β[π] = (0.99)π π’[π + 3]
3. β[π] = (4)−π π’[2 − π]
A continuous – time LTI system is characterized by impulse response. Determine whether
the impulse response is (a) Stable and (b) causal.
1. β(π‘) = π −3π‘ π’(π‘ − 1)
2. β(π‘) = π −π‘ π’(π‘ + 100)
3. β(π‘) = π π‘ π’(−π‘ − 1)
4. β(π‘) = π −4|π‘|
Find the step response for the LTI system represented by the impulse response
1 π
1. β[π] = (2) π’[π]
2. β(π‘) = π‘ π’(π‘)
3. β(π‘) = π −|π‘|
4. β(π‘) = πΏ(π‘) − πΏ(π‘ − 1)
Determine convolution of the two given sequences
1. π₯[π] = {1, β
2 , 3, 4} πππ β[π] = {1
β , 1, 3, 2}
2. π₯[π] = {1
β , 2, 4} πππ β[π] = {1
β , 1, 1, 1, 1, 1}
Sketch the direct form I and direct form II implementations for the following
1
1. π¦[π] + 2 π¦[π − 1] − π¦[π − 3] = 3π₯[π − 1] + 2π₯[π − 2]
1
1
1
2. π¦[π] − 4 π¦[π − 1] + 8 π¦[π − 2] = π₯[π] + 2 π₯[π − 2]
3.
π 3 π¦(π‘)
ππ‘
+2
π 2 π¦(π‘)
ππ‘
+ 3π¦(π‘) = π₯(π‘) + 3
ππ₯(π‘)
ππ‘
Solve the following homogeneous differential and difference equations with the specified
initial conditions.
1.
2.
3.
π 2 π¦(π‘)
ππ‘
π 2 π¦(π‘)
ππ‘
π 3 π¦(π‘)
ππ‘
+3
ππ¦(π‘)
+2
ππ¦(π‘)
+
ππ‘
ππ‘
π 2 π¦(π‘)
ππ‘
+ 2π¦(π‘) = 0
+ π¦(π‘) = 0
−
ππ¦(π‘)
ππ‘
;
;
π¦(0) = 0 ;
π¦(0) = 1 ;
π¦ ′ (0) = 2
π¦ ′ (0) = 1
− π¦(π‘) = 0 ; π¦(0) = 1 ; π¦ ′ (0) = 1; π¦ ′′ (0) = −2
4. π¦[π] − 3π¦[π − 1] − 4π¦[π − 2] = 0