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A continuous time x(t) signal is given below. x(t) =

{
t+2, 0 ≤ t < 2,
6−t, 2 ≤ t < 4,
2, 4 ≤ t ≤ 5
}

a) x(t−2)
b) x(−t)
c) x(−t−2)
d) x(t/2)
e) x(t).δ(t−4.5)
f) ∫ from -[infinity] to [infinity] x(t)δ(t−4.5)dt
g) x(t)⋅u(t−2.5)
h) x(t)⋅[u(t−1)−u(t−3)]

İşaretlerinin grafiğini çiziniz.

User Kajuan
by
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1 Answer

4 votes

Final answer:

To find the signal for each given expression, we need to perform the specified operations on the given continuous time signal. (a) x(t-2) involves shifting the signal 2 units to the right. (b) x(-t) involves reflecting the signal about the y-axis. (c) x(-t-2) involves reflecting the signal about the y-axis and shifting it 2 units to the left. (d) x(t/2) involves stretching the signal horizontally by a factor of 2. (e) x(t) is the given signal. (f) x(t).δ(t-4.5) involves multiplying the signal by the Dirac delta function shifted 4.5 units to the right. (g) x(t)⋅u(t-2.5) involves multiplying the signal by the unit step function shifted 2.5 units to the right. (h) x(t)⋅[u(t-1)-u(t-3)] involves subtracting two unit step functions shifted 1 unit to the right and 3 units to the right, respectively.

Step-by-step explanation:

To find the signal for each given expression:

a) To find x(t-2), we shift the signal 2 units to the right. Since 0 <= t < 2, the shifted signal will be t+2-2 = t. When 2 <= t < 4, the shifted signal will be 6-t-2 = 4-t. And when 4 <= t <= 5, the shifted signal will be 2.

b) To find x(-t), we reflect the signal about the y-axis. Since 0 <= t < 2, the reflected signal will be -(t+2). When 2 <= t < 4, the reflected signal will be -(6-t). And when 4 <= t <= 5, the reflected signal will be -(2).

c) To find x(-t-2), we reflect the signal about the y-axis and shift it 2 units to the left. Since 0 <= t < 2, the resulting signal will be -(t+2+2) = -t-4. When 2 <= t < 4, the resulting signal will be -(6-t+2) = -4+t. And when 4 <= t <= 5, the resulting signal will be -(2).

d) To find x(t/2), we stretch the signal horizontally by a factor of 2. Since 0 <= t < 2, the stretched signal will be (t/2)+2. When 2 <= t < 4, the stretched signal will be (6-t)/2 = 3-(t/2). And when 4 <= t <= 5, the stretched signal will be 2.

e) To find x(t), we don't need to do any operations on the signal since it is already x(t).

f) To find x(t).δ(t-4.5), we multiply the signal by the Dirac delta function shifted 4.5 units to the right. The resulting signal will be x(t) when t = 4.5, and 0 otherwise.

g) To find x(t)⋅u(t-2.5), we multiply the signal by the unit step function shifted 2.5 units to the right. Since the unit step function is 0 for t < 2.5 and 1 for t >= 2.5, the resulting signal will be x(t) for t >= 2.5, and 0 otherwise.

h) To find x(t)⋅[u(t-1)-u(t-3)], we subtract two unit step functions shifted 1 unit to the right and 3 units to the right, respectively. Since the first unit step function is 0 for t < 1 and 1 for t >= 1, and the second unit step function is 0 for t < 3 and 1 for t >= 3, the resulting signal will be x(t) for 1 <= t < 3, and 0 otherwise.

User Rob Monhemius
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