ECE 350 HW #1 Name Fall 2013 - Find The Zero-State Response

ECE 350 – HW #1 Name Fall 2013– Find the zero-state response of a system with impulse response h(t)

Evaluate the zero-state response of a linear time-invariant system given its impulse response and input signal. Specifically, for a system with impulse response h(t) = t[u(t) – u(t – 1)] and input x(t) = 2[u(t) – u(t – 2)].

The zero-state response of a system is the convolution of the input signal with the system's impulse response. Mathematically, it is expressed as:

y_zs(t) = (x * h)(t) = ∫_-∞^∞ x(τ) h(t – τ) dτ.

Given the functions are multiplied by unit step functions, their effective support is limited to specific intervals, simplifying the convolution integral to finite bounds.

Solution

First, note the input and impulse response functions:

• Input: x(t) = 2[u(t) – u(t – 2)]
• Impulse response: h(t) = t[u(t) – u(t – 1)]

Both functions are non-zero only within finite intervals:

• x(t) is non-zero between 0 and 2.
• h(t) is non-zero between 0 and 1.

Therefore, the convolution integral simplifies to:

y_zs(t) = ∫₀^2 x(τ) h(t – τ) dτ.

Because x(τ) = 2 for τ in [0, 2] and zero elsewhere, substitute and set up the integral accordingly:

y_zs(t) = 2 ∫₀^2 h(t – τ) dτ.

Now, examine h(t – τ):

• h(t – τ) = (t – τ) [u(t – τ) – u(t – τ – 1)].

Since u(t – τ) and u(t – τ – 1) are step functions, the support of h(t – τ) depends on the value of t.

To evaluate the convolution, consider different ranges of t where h(t – τ) is non-zero.

Case 1: t

For t t.

Since τ ≥ 0 and t

Case 2: 0 ≤ t ≤ 1

In this case, t – τ ≥ 0 for τ ≤ t; and t – τ – 1 t – 1, which is less than or equal to t. Since t ≤ 1, the interval where h(t – τ) is non-zero is between τ = 0 and τ = t, because for τ in [0, t], t – τ ≥ 0, and t – τ – 1

Thus, for τ in [0, t], h(t – τ) = (t – τ). The convolution becomes:

 yzs(t) = 2 ∫0^{t} (t – τ) dτ.

Compute the integral:

 yzs(t) = 2 [ tτ – (1/2) τ2 ] from τ=0 to τ=t = 2 [ t  t – (1/2) t2 ] = 2 [ t2 – (1/2) t2 ] = 2  (1/2) t2 = t2.

Case 3: 1

In this interval, the support of h(t – τ) is restricted to τ in [t – 1, t] because between these bounds, t – τ ≥ 0 and t – τ – 1 ≥ 0.

Specifically, h(t – τ) is non-zero when t – τ ≥ 0 and t – τ – 1 ≥ 0, i.e., τ in [t – 1, t].

Since the integral is over [0, 2], and τ in [t – 1, t], intersected with [0, 2], the limits are from max(0, t – 1) to min(2, t).

Because t ≤ 2, t – 1 ≥ 0 for t ≥ 1.

Thus, for 1

 yzs(t) = 2 ∫t−1^{t} (t – τ) dτ.

Evaluate this integral:

 yzs(t) = 2 [ tτ – (1/2) τ2 ] from τ= t−1 to τ= t.

Calculate at the bounds:

At τ= t: t * t – (1/2) t2 = t2 – (1/2) t2 = (1/2) t2.

At τ= t−1: t * (t−1) – (1/2) (t−1)2 = t2 – t – (1/2)(t2 – 2t + 1) = t2 – t – (1/2)t2 + t – 1/2 = (1/2)t2 – 1/2.

Subtract the two:

 yzs(t) = 2 [ (1/2)t2 – ( (1/2)t2 – 1/2) ] = 2 [ (1/2)t2 – 1/2 t2 + 1/2 ] = 2 * (1/2) = 1.

Thus, for t in (1, 2], the zero-state response is constant at 1.

Case 4: t > 2

For t > 2, the limits of integration are from τ = t – 1 to τ = t, but since τ only goes up to 2 because of the support of x(τ) in [0, 2], and t – 1 ≥ 1, the upper limit is t, which exceeds 2. But since x(τ) is zero outside [0, 2], the integral reduces to τ in [0, 2], with h(t – τ) non-zero only if t – τ between 0 and 1. Because t – τ ≥ 0, and τ in [0, 2], the limits are effectively from τ=0 to τ=2.

Given the support and the finite duration, for t ≥ 3, the overlap shrinks to an empty set, and the response is zero.

Final expression of the zero-state response y_zs(t):

• For t zs(t) = 0.
• For 0 ≤ t ≤ 1: y_zs(t) = t².
• For 1 zs(t) = 1.
• For t > 2: y_zs(t) = 0.

Final answer:

The zero-state response of the given system to the specified input is

y_zs(t) =

• 0 for t < 0
• t² for 0 ≤ t ≤ 1
• 1 for 1 < t ≤ 2
• 0 for t > 2

References

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