(Note: here are the solution, only showing you the approach to solve the
problems. If you find some typos or calculation error, please post it on Piazza
and let us know )
1.17 : Consider a continuous-time system with input x(t) and output y(t)
related by y(t) = x( sin(t)).
(a) Is this system Casual?
The system is not causal because for example the output at t= -π = -3.14, depends
on the input at t= 0; In other words:
y(-π) = x(sin(-π)) = x(0);
So, the output of the system at t = -π depends on the input at t = 0 which the future
time.
Thus, the system is not causal.
(b) Is this system linear?
For input x1(t), we have y1(t) = x1(sin(t)), and
For input x2(t), we have y2(t) = x2(sin(t)), and,
For input x3(t)=a x1(t)+b x2(t), we have
y3(t) = ax1(sin(t)) + bx2(sin(t))
= a y1(t)+b y2(t);
Thus, the system is linear.
1.19 a: determine whether the system is linear, time invariant or both.
y(t)=t2x(t-1)
Linearity:
For input x1(t), we have y1(t) = t2 x1(t-1), and
For input x2(t), we have y2(t) = t2x2(t-1), and,
For input x3(t)=a x1(t)+b x2(t), we have
y3(t) = t2 x3(t-1)
=t2 [ax1(t-1)) + bx2(t-1)]
= a t2 x1(t-1)+ b t2 x2(t-1)
= a y1(t)+b y2(t);
Thus, the system, is linear.
Time Variant:
Let’s assume x1(t)=x(t-to), so, for this input we have the output as;
y1(t)= t2 x1(t-1)
if we substitute x1(t-1)=x(t-1-to), we get:
y1(t)= t2 x(t-1-to)
On the other hand, let’s compute y(t-to) as;
y(t-to)= (t-to)2 x(t-to-1)
Since y1(t)≠ y(t-to), thus, the system is not time invariant, In other words it is time
variant.
1.27 b: Consider the system of y(t)=[cos(3t)]x(t);
(1) memoryless?!
The system is memoryless, because at any t, y(t) depends on the input at exactly
time t, i.e., x(t). In other words, the output doesn’t depend on the past or future of
the input.
(2) Time Invariant?!
Let’s assume x1(t)=x(t-to), so, for this input we have the output as;
y1(t)= [cos(3t)] x1(t)
if we substitute x1(t)=x(t-to), we get:
y1(t)= [cos(3t)] x(t-to)
On the other hand, let’s compute y(t-to) as;
y(t-to)= [cos(3(t-to))] x(t-to)
Since y1(t)≠ y(t-to), thus, the system is not time invariant, In other words it is time
variant.
(3) Linear?!
For input x1(t), we have y1(t) = [cos(3t)] x1(t), and
For input x2(t), we have y2(t) = [cos(3t)]x2(t), and,
For input x3(t)=a x1(t)+b x2(t), we have
y3(t) = [cos(3t)] x3(t)
=[cos(3t)] [ax1(t)) + bx2(t)]
= a [cos(3t)] x1(t)+ b [cos(3t)] x2(t)
= a y1(t)+b y2(t);
Thus, the system, is linear.
(4) Causal?!
The system is memoryless, because at any t, y(t) depends on the input at exactly
time t, i.e., x(t). In other words, the output doesn’t depend on the future of the input.
(5) Stable?!
The system is stable because;
Let’s assume that |x(t)|<B, which B is a finite number,
Since we have |cos(3(t)|<= 1, so, we get;
|y(t)|=|cos(3t)|. |x(t)| < M
In other words, if the input is bounded, the output will be bounded. Thus the
system is stable.
1. 30 a and b: Invertibility and inverse system
A system is said to be invertible if distinct inputs lead to distinct output. Here are
the solution for (a) and (b):
(a) invertible, the inverse system is w(t) = y(t + 4).
(b) not invertible, you can plot the cos⁡(x(t)), you will find that two inputs x(t),
π
π
may correspond to the same output y(t), for example: and 5 , will both get 1/2.
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2.22 b and d: use the convolution integral to find 𝐲(𝐭)
(b) Here is the definition of convolution:
∞
∞
y(t) = x(t) ∗ h(t) = ∫ x(τ) h(t − τ)dτ = ∫ x(t − τ) h(τ)dτ
−∞
−∞
For (b), the equation becomes:
∞
y(t) = ∫ x(τ) 𝑒 2𝑡−2τ u(1 − (t − τ))dτ
−∞
∞
= ∫−∞ x(τ) 𝑒 2𝑡−2τ u(1 − t + τ)dτ
∞
⁡⁡⁡⁡⁡= ∫ x(τ) 𝑒 2𝑡−2τ dτ
t−1
The last step is obtained through
1 − t + τ ≥ 0⁡⁡thus⁡τ ≥ t − 1
From the equation of x(t), we know that
x(t) = 1⁡if⁡0 ≤ t < 2, x(t) = −1⁡if⁡2 < t ≤ 5, and⁡x(t) = 0⁡if⁡otherwise
*here is the sketch of x(t)*
1
0
5
2
-1
Given these constraints, the above integral can be simplified into
∞
y(t) = ∫ x(τ) 𝑒
2𝑡−2τ
dτ = 𝑒
2𝑡
∞
∫ x(τ) 𝑒 −2τ dτ
t−1
t−1
Given the bound of x(t), there are four cases in solving this integral:
(1) If t − 1 < 0⁡(t < 1), as illustrated in the following figure, then
1
t-1
0
5
2
-1
y(t) = 𝑒
2𝑡
∞
∫ x(τ) 𝑒
−2τ
dτ = 𝑒
2𝑡
t−1
2
(∫ 1 ∗ 𝑒
−2τ
5
dτ + ∫ (−1) ∗ 𝑒 −2τ dτ)
0
2
1
1
2
2
= −𝑒 2𝑡 (𝑒 −4 − 𝑒 −10 − )
(2) If⁡0 < t − 1 < 2⁡(1 < t < 3), as illustrated in the following figure, then
1
0
-1
t-1
2
5
y(t) = 𝑒
2𝑡
∞
∫ x(τ) 𝑒
−2τ
2
2𝑡
dτ = 𝑒 (∫ 1 ∗ 𝑒
t−1
−2τ
5
dτ + ∫ (−1) ∗ 𝑒 −2τ dτ)
t−1
2
1
1
2
2
= −𝑒 2𝑡 (𝑒 −4 − 𝑒 −10 − 𝑒 −2𝑡+2 )
(3) If⁡2 < t − 1 < 5⁡(3 < t < 6), as illustrated in the following figure, then
1
0
2
5
t-1
-1
2𝑡
∞
y(t) = 𝑒 ∫ x(τ) 𝑒
−2τ
dτ = 𝑒
2𝑡
5
(∫ −1 ∗ 𝑒 −2τ dτ)
t−1
t−1
1
1
2
2
= −𝑒 2𝑡 ( 𝑒 −2𝑡+2 − 𝑒 −10 )
(4) If⁡5 < t − 1⁡(t > 6), as illustrated in the following figure, then
1
0
5
2
t-1
-1
y(t) = 𝑒
2𝑡
∞
∫ x(τ) 𝑒
−2τ
t−1
2𝑡
∞
dτ = 𝑒 (∫ 0 ∗ 𝑒 −2τ dτ)
t−1
=0
From the above cases, you can get the y(t) and do the sketch.
(d) The approach in (b) is a general, and could be applied in most of the cases.
First from the sketch in Fig. P2.22, we know that x(t) = a ∗ t + b.
∞
∞
y(t) = x(t) ∗ h(t) = ∫ x(τ) h(t − τ)dτ = ∫ x(t − τ) h(τ)dτ
−∞
−∞
For (d), the equation becomes (we use the second half of the above equation, for
simplification consideration)
∞
y(t) = ∫ h(τ) (at + b − aτ)dτ
−∞
∞
∞
= (a ∗ t + b) ∫−∞ h(τ) dτ + ⁡a ∫−∞ τ ∗ h(τ) dτ
The first half of above equation is easy to calculate, we can get
∞
∫ h(τ) dτ = 1
−∞
Then we get:
∞
y(t) = ∫ h(τ) (at + b − aτ)dτ
−∞
∞
= a ∗ t + b + ⁡a ∫−∞ τ ∗ h(τ) dτ
1
4
∞
1
= a ∗ t + b + ⁡a(∫0 τ ∗ dτ + ∫1 −τ ∗ δ(τ − 2)dτ)
3
2
2
3
3
= a ∗ t + b + ⁡a( − ) = a ∗ t + b
Then you can get the y(t) and do the sketch.
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