Signal Transformation
The signal \( x\!\left(\frac{t-1}{2}\right) \) is a transformed version of \( x(t) \). It differs from \( x(t) \) in two ways:
- Time scaling (stretching)
- Time shifting (delay)
1. Compare the Arguments
Original signal:
\[ x(t) \]
Transformed signal:
\[ x\!\left(\frac{t-1}{2}\right) \]
The key expression is:
\[ \frac{t-1}{2} \]
2. Rewrite the Expression
\[ \frac{t-1}{2} = \frac{1}{2}(t-1) \]
Step A: Time Scaling (Stretching)
The factor 1/2 inside the argument causes time expansion.
- If \( |a| > 1 \) then the signal compresses
- If \( 0 < |a| < 1 \) then the signal stretches
Since \( a = 1/2 \), the signal is:
Stretched by factor 2
Step B: Time Shift (Delay)
To find the shift, set:
\[ \frac{t-1}{2} = 0 \]
Then:
\[ t = 1 \]
The signal is delayed by 1 unit.
Final Interpretation
- Stretched by factor 2
- Shifted right (delayed) by 1
Example Using a Rectangular Pulse
Assume:
\[ x(t) = \begin{cases} 1 & 0 \le t \le 2 \\ 0 & \text{otherwise} \end{cases} \]
Original Signal Diagram
Amplitude
1 | ┌──────────┐
| │ │
0 |────────┘ └────────────
0 2 t
After Stretching: \( x(t/2) \)
Solve:
\[ 0 \le \frac{t}{2} \le 2 \]
\[ 0 \le t \le 4 \]
Amplitude
1 | ┌──────────────────┐
| │ │
0 |────────┘ └────────────
0 4 t
After Stretch + Shift: \( x\left(\frac{t-1}{2}\right) \)
Solve:
\[ 0 \le \frac{t-1}{2} \le 2 \]
\[ 1 \le t \le 5 \]
Amplitude
1 | ┌──────────────────┐
| │ │
0 |────────────┘ └────────────
0 1 5 t
Visual Comparison
| Signal | Starts At | Ends At | Effect |
|---|---|---|---|
| \( x(t) \) | 0 | 2 | Original |
| \( x(t/2) \) | 0 | 4 | Stretched |
| \( x\left(\frac{t-1}{2}\right) \) | 1 | 5 | Stretched + Shifted Right |
Intuition Trick
- Set the inside expression equal to 0.
- Solve for \( t \).
- That tells you where features move.