1-Mark Questions
QQuestion 1 1 Mark
A Linear Time Invariant system with an impulse response \(h(t)\) produces output \(y(t)\) when input \(x(t)\) is applied. When the input \(x(t - \tau)\) is applied to a system with impulse response \(h(t - \tau)\), the output will be
AOptions
- \(y(t)\)
- \(y(2(t - \tau))\)
- \(y(t - \tau)\)
- \(y(t - 2\tau)\)
SSolution
LTI System properties:
Time-Invariance property:
If input \(x(t)\) produces output \(y(t)\), then:
Given system:
Original: \(x(t) * h(t) = y(t)\)
Modified system:
- Input: \(x(t - \tau)\) (delayed input)
- Impulse response: \(h(t - \tau)\) (delayed system)
Output calculation:
Using convolution properties:
Let \(\beta = \alpha - \tau\):
But we know that \(x(t) * h(t) = y(t)\)
By time-invariance:
Alternative reasoning:
Step 1: Delay input by \(\tau\)
Step 2: Delay system by \(\tau\)
Step 3: Both delayed by \(\tau\)
The total delay is \(\tau + \tau = 2\tau\).
Correct answer: D
2-Mark Questions
QQuestion 2 2 Mark
The circuit shows an ideal diode connected to a pure inductor and is connected to a purely sinusoidal 50 Hz voltage source. \(V_s = 10\sin(100\pi t)\), \(L = (0.1/\pi)\) H. Under ideal conditions, the current waveform through the inductor will look like:
SSolution
Given:
- \(V_s = 10\sin(100\pi t)\) V
- \(\omega = 100\pi\) rad/s
- \(L = 0.1/\pi\) H
- Ideal diode in series
Inductive reactance:
Circuit behavior:
Positive half-cycle (\(V_s > 0\)):
- Diode forward biased (conducts)
- Current flows through inductor
- \(V_L = L\frac{di}{dt} = V_s\)
- \(i = \frac{1}{L}\int V_s dt = \frac{1}{L} \int 10\sin(100\pi t)dt\)
- \(i = \frac{10}{L \times 100\pi}[-\cos(100\pi t)]\)
- \(i = \frac{10 \pi}{0.1 \times 100\pi}[-\cos(100\pi t)] = -\cos(100\pi t) + C\)
Negative half-cycle (\(V_s < 0\)):
- Diode reverse biased (blocks)
- Current cannot flow (ideal diode)
- But inductor cannot have instantaneous current change!
Key insight:
When diode tries to block (at \(\omega t = \pi\)), the inductor current cannot suddenly go to zero. The inductor will try to maintain current, but diode blocks reverse current.
Result:
- Current flows during positive half of \(V_s\)
- Current is sinusoidal (lagging by 90° in pure L)
- During negative half, current forced to zero by diode
- Creates half-wave rectified sine wave pattern
Peak current:
The waveform shows:
- Sinusoidal rise from 0 to 1A
- Sinusoidal fall from 1A to 0
- Zero during negative half
- Period = 20ms (50Hz)
Correct answer: Waveform showing half-wave rectified current with peak of 1A
QQuestion 3 2 Mark
In the chopper circuit shown, the main thyristor (TM) is operated at a duty ratio of 0.8 which is much larger than the commutation interval. If the maximum allowable reapplied dv/dt on TM is 50 V/\(\mu\)s, what should be the theoretical minimum value of \(C_1\)? Assume current ripple through \(L_o\) to be negligible.
AOptions
- 0.2 \(\mu\)F
- 2 \(\mu\)F
- 0.02 \(\mu\)F
- 20 \(\mu\)F
SSolution
Given:
- Source voltage: \(V_s = 100\) V
- Duty ratio: \(D = 0.8\)
- Maximum dv/dt: \(50\) V/\(\mu\)s
- Load current ripple negligible (constant \(I_o\))
Commutation analysis:
During commutation:
When auxiliary thyristor fires to turn off main thyristor:
- Capacitor voltage reverses and applies across TM
- Rate of voltage rise across TM must be limited
dv/dt calculation:
The capacitor discharges through load current:
where \(I\) is the load current.
For constant load current \(I_o\):
Maximum dv/dt constraint:
Load current estimation:
Output voltage: \(V_o = D \times V_s = 0.8 \times 100 = 80\) V
Without load resistance value, assuming typical chopper operation:
If load resistance \(R_L\) is given or can be inferred:
From answer choices and typical values:
Correct answer: B