Questions: Basic Signal Operations and Transformations

Question 1

A signal x(t) has a peak at t = 0. Where does the peak of y(t) = x(2t − 4) occur?

• A. t = 4, because the −4 term shifts the signal right by 4
• B. t = 2, because the argument equals zero when 2t − 4 = 0, so t = 2
• C. t = −2, because the signal is compressed by 2 and shifted left
• D. t = 8, because the shift is amplified by the time-scaling factor

Answer: (B)

The peak of x(t) occurs when its argument equals zero. For y(t) = x(2t − 4), set 2t − 4 = 0, giving t = 2. The effective shift is 4/2 = 2, not 4. This is the critical point: x(at − b) should be rewritten as x(a(t − b/a)), so the time shift is b/a, not b. Option A is the classic mistake of reading the shift directly from the expression without accounting for the time-scaling factor. Option D doubles the shift rather than dividing.

Question 2

A signal x(t) represents a speech waveform. What does the transformed signal y(t) = x(0.5t) sound like compared to x(t)?

• A. It sounds sped up — the speech is compressed in time and perceived as faster
• B. It sounds slowed down — the speech is stretched in time and perceived as slower
• C. It sounds identical but quieter — amplitude scaling does not affect timing
• D. It sounds pitch-shifted downward without changing the playback speed

Answer: (B)

Time scaling with a = 0.5 (less than 1) stretches the signal: x(0.5t) takes twice as long to complete each feature, making the speech play back at half speed and sound slowed down. Compression (a > 1) would speed it up. Option A describes the a > 1 case. Option C confuses amplitude scaling with time scaling. Option D is an approximation of pitch shifting, but x(0.5t) fundamentally slows the signal, which does lower pitch as a side effect — yet describing it as 'pitch shifted without speed change' is incorrect.

Question 3

In the signal x(t − t₀), a positive value of t₀ shifts the signal to the right (delay) along the time axis.

• T. True
• F. False

Answer: True

A positive t₀ in x(t − t₀) delays the signal — the event that occurred at t = 0 in x(t) now occurs at t = t₀ in the shifted version. Graphically, the signal moves to the right. This direction is counterintuitive to many students because subtracting inside the function (−t₀) seems like it should shift left. The rule to remember: subtract from t to delay (shift right); add to t to advance (shift left). This is consistent with the general x(a(t − t₀)) form: t₀ is the time shift, and it is to the right when positive.

Question 4

In the expression x(2t − 6), the time shift of the original signal x(t) is 6 units to the right.

• T. True
• F. False

Answer: False

The shift is 3 units, not 6. To find the true shift, rewrite the expression in standard form: x(2t − 6) = x(2(t − 3)). In the form x(a(t − t₀)), the time shift t₀ = 3 and the scale factor a = 2. Reading the shift directly from the coefficient (−6 ÷ 1 = 6) without accounting for the scale factor (dividing by a = 2) is the most common error. The effective shift is always b/a when the expression is written as x(at − b).

Question 5

A signal is given as y(t) = x(3t + 6). Describe step-by-step how y(t) relates to x(t) — what transformations have been applied and in what order?

Model answer: Rewrite: x(3t + 6) = x(3(t + 2)). This reveals two transformations: (1) time shift by −2 (since the form is t − t₀ with t₀ = −2, the signal is advanced — shifted LEFT by 2 units); (2) time compression by factor 3 (the signal runs through its pattern 3× faster). Applying in the natural order: first shift x(t) left by 2 to get x(t + 2), then compress by 3 to get x(3t + 6). The resulting signal is both earlier (shifted left) and faster (compressed).

The factoring step — rewriting x(at + b) as x(a(t + b/a)) — is the key mechanical skill. Students who apply shifts and scales without factoring first will consistently get the wrong effective shift when a ≠ 1. The question also tests understanding of direction: adding inside the argument shifts left (advances), subtracting shifts right (delays).