Kit Library / Engineering / Signals and Systems
⚡ Topic Learning KitShared by a Veda learner · Generated with Veda AI
Veda Bites are swipeable micro-lessons — each one teaches exactly one idea. The bites below come from this kit — swipe through them in the app.
💡 Key Idea
Shift, flip, and scale systematically.
To sketch transformed signals like $x(at+b)$, always factor the argument first: $x(a(t + b/a))$. Then apply operations in this order: scale (multiply $t$ by $a$), shift (add/subtract $b/a$), and flip (if $a$ is negative). This prevents common sketching errors.
↳ Factor the argument first; then scale, shift, and flip in that order.
📖 Definition
The on/off switch for signals.
The continuous-time unit step $u(t)$ is 0 for $t<0$ and 1 for $t\geq 0$. The discrete-time step $u[n]$ is 0 for $n<0$ and 1 for $n\geq 0$. It is used to gate or truncate other signals.
↳ The step function turns signals on at a specified time.
📖 Definition
The step's integral — a linear rise.
The unit ramp $r(t)$ equals $t$ for $t\geq 0$ and 0 for $t<0$. It is the integral of the step function. Ramp signals increase linearly with time after they turn on.
↳ A ramp is a linearly increasing signal that starts at zero.
💡 Key Idea
Find where the argument crosses zero.
To sketch $u(-t+4)$, find the zero crossing of the argument: $-t+4=0 \Rightarrow t=4$. Since the coefficient of $t$ is negative, the step is flipped. So $u(-t+4)$ is 1 for $t<4$ and 0 for $t\geq 4$.
↳ Set the step argument to zero to find the transition point.
💡 Key Idea
A window of width Π.
The rectangular pulse $\Pi(t)$ is 1 for $|t| < 1/2$ and 0 otherwise. The expression $\Pi((t-2)/4)$ represents a pulse centered at $t=2$ with width 4. Adding $\Pi(t-2)$ gives a pulse centered at 2 with width 1.
↳ Pulse width and center are read directly from the argument form.
💡 Key Idea
Truncating a ramp with a step.
The signal $r(t) \cdot u(-t+3)$ is the ramp $r(t)$ multiplied by a step that is 1 for $t<3$. This means the ramp is active only for $t<3$. For $t<0$, $r(t)=0$, so the signal is 0. For $0 \leq t < 3$, it equals $t$. For $t\geq 3$, it is 0.
↳ Multiplying by a flipped step truncates the ramp at the step's edge.
💡 Key Idea
Sum shifted ramps to build shapes.
The signal $r(t) - 2r(t-2) + r(t-4)$ creates a triangular shape. For $0 \leq t < 2$: only $r(t)$ is active → slope 1. For $2 \leq t < 4$: $r(t) - 2r(t-2)$ → slope -1. For $t \geq 4$: all terms cancel → value 0.
↳ Weighted ramp sums create piecewise-linear signals with controlled slopes.
💡 Key Idea
Difference of steps creates a window.
For $x[n] = u[-n+2] - u[-n-2]$: The first step is 1 for $n \leq 2$. The second is 1 for $n \leq -2$. Subtracting gives 1 only where the first is 1 and the second is 0, i.e., for $-1 \leq n \leq 2$.
↳ Subtracting two flipped steps yields a finite discrete pulse.
📖 Definition
Repeats exactly after a fixed time.
A continuous-time signal $x(t)$ is periodic if there exists a positive $T_0$ such that $x(t + T_0) = x(t)$ for all $t$. The smallest such $T_0$ is the fundamental period.
↳ Periodicity means exact repetition after the fundamental period $T_0$.
➗ Formula
Frequency determines the period.
For $x(t) = \cos(\omega_0 t)$ or $\sin(\omega_0 t)$, the fundamental period is $T_0 = 2\pi/\omega_0$. For $\cos(6\pi t)$, $\omega_0 = 6\pi$, so $T_0 = 2\pi/(6\pi) = 1/3$ seconds.
↳ Period is $2\pi$ divided by the angular frequency.
💡 Key Idea
Euler's formula links to cos/sin.
For $x(t) = e^{j\omega_0 t}$, the fundamental period is also $T_0 = 2\pi/|\omega_0|$. For $e^{j8\pi t}$, $\omega_0 = 8\pi$, so $T_0 = 2\pi/(8\pi) = 1/4$ seconds.
↳ Complex exponentials are periodic with period $2\pi/|\omega_0|$.
💡 Key Idea
Find the LCM of individual periods.
For $x(t) = 2\cos(50\pi t) + 3\sin(25\pi t)$: First term has $T_1 = 2\pi/(50\pi) = 1/25$. Second has $T_2 = 2\pi/(25\pi) = 2/25$. The fundamental period is the least common multiple (LCM) of $1/25$ and $2/25$, which is $2/25$ seconds.
↳ The sum is periodic if the ratio of periods is rational; period is their LCM.
💡 Key Idea
Constants and steps break periodicity.
The signal $x(t) = 2u(t) + 3\cos(2\pi t)$ is not periodic because the step $u(t)$ is not periodic. Adding a non-periodic component to a periodic one makes the entire signal non-periodic.
↳ Any signal containing a non-periodic component like a step is non-periodic.
💡 Key Idea
Compare frequencies to find the period.
For $x(t) = 3e^{j(4\pi t + \pi/3)} + 4e^{-j(2t + \pi/2)}$: First term has $\omega_1 = 4\pi$, so $T_1 = 1/2$. Second has $\omega_2 = 2$, so $T_2 = \pi$. Since $T_1/T_2 = (1/2)/\pi = 1/(2\pi)$ is irrational, the sum is not periodic.
↳ If the ratio of individual periods is irrational, the sum is non-periodic.
💡 Key Idea
Constants are periodic with any period.
For $x(t) = 2 + \sin(4\pi t)$: The constant 2 is periodic with any period. The sine term has $T = 2\pi/(4\pi) = 1/2$. Therefore, the sum is periodic with fundamental period $T_0 = 1/2$ seconds.
↳ Adding a constant to a periodic signal does not change its fundamental period.
🎯 Exam Tip
Rational ratio means periodic.
↳ Test rationality of period ratios to determine periodicity.
🧠 Memory Trick
SSF order prevents sketching errors.
When transforming $x(t)$ to $x(at+b)$, always remember SSF: first Scale, then Shift, then Flip (if needed). This order guarantees correct sketches every time.
↳ Scale, Shift, Flip — in that order — for accurate signal sketches.
❓ Did You Know
Periodic signals enable Fourier analysis.
Periodic signals can be represented as sums of sinusoids (Fourier series). This is fundamental to analyzing LTI systems, filtering, and communications. Knowing the fundamental period is the first step.
↳ Periodicity is the gateway to frequency-domain analysis.
This topic covers sketching basic continuous-time and discrete-time signals (step, ramp, pulse) under transformations like shifting, scaling, and flipping. It also covers determining periodicity and fundamental periods f…
Full explanation, worked examples & memory tricks in the app
All 15 questions, laddered from warm-up to topper-level, each with its answer and the reasoning behind it.
Q1. For the signal $x(t) = u(-t + 4)$, what is the value of $x(t)$ for $t > 4$?
BeginnerB · 0
The argument $-t+4$ is negative for $t > 4$. The unit step $u(\text{negative}) = 0$, so $x(t)=0$ for $t > 4$.
Q2. The signal $x(t) = \Pi\left(\frac{t-2}{4}\right)$ is a rectangular pulse. What is its time duration?
BeginnerC · 4 seconds
The standard rectangular pulse $\Pi(t)$ has duration 1. Scaling the argument by $\frac{1}{4}$ stretches the duration by a factor of 4, so the duration is 4 seconds.
Q3. Given $x(t)$ is a triangular pulse with peak amplitude 2 at $t=0$ and zero outside $[-1, 1]$, what is the peak amplitude of $3x(5t)$?
BeginnerA · 6
Amplitude scaling by 3 multiplies the peak amplitude by 3. The original peak is 2, so the new peak is $3 \times 2 = 6$.
Q4. What is the fundamental period $T_0$ of $x(t) = \cos(6\pi t)$?
BeginnerD · $\frac{1}{3}$ seconds
For $\cos(\omega_0 t)$, the fundamental period is $T_0 = \frac{2\pi}{\omega_0}$. Here $\omega_0 = 6\pi$, so $T_0 = \frac{2\pi}{6\pi} = \frac{1}{3}$ seconds.
Q5. Which signal expression correctly represents a ramp that starts at $t=4$ and has slope 1?
BeginnerC · $r(t-4)$
The unit ramp $r(t)$ is zero for $t
Q6. The signal $x(t) = 2 + \sin(4\pi t)$ is periodic with fundamental period $T_0 = \frac{1}{2}$ seconds.
BeginnerA · True
The constant 2 does not affect periodicity. The sine term has frequency $4\pi$ rad/s, so $T_0 = \frac{2\pi}{4\pi} = \frac{1}{2}$ seconds.
Q7. Which of the following is the correct expression for the signal $x(t) = r(t) - 2r(t-2) + r(t-4)$ for $2 \le t < 4$?
IntermediateB · $x(t) = -t + 4$
For $2 \le t < 4$, $r(t) = t$ and $r(t-2) = t-2$, while $r(t-4) = 0$. So $x(t) = t - 2(t-2) = t - 2t + 4 = -t + 4$.
Q8. For the signal $x(t)$ (triangular pulse, peak 2 at $t=0$, zero outside $[-1,1]$), what is the value of $x(2t+2)$ at $t = -1$?
IntermediateD · 2
At $t = -1$, the argument is $2(-1) + 2 = 0$. Since $x(0) = 2$ (peak), $x(2t+2)$ at $t=-1$ equals 2.
Q9. Which of the following signals is NOT periodic?
IntermediateA · $x(t) = 2u(t) + 3\cos(2\pi t)$
The unit step $u(t)$ is not periodic (it is constant for $t>0$ and zero for $t
Q10. The fundamental period of $x(t) = 2\cos(50\pi t) + 3\sin(25\pi t)$ is:
IntermediateC · $\frac{2}{25}$ seconds
The period of $\cos(50\pi t)$ is $\frac{2\pi}{50\pi} = \frac{1}{25}$ s. The period of $\sin(25\pi t)$ is $\frac{2\pi}{25\pi} = \frac{2}{25}$ s. The fundamental period is the LCM of $\frac{1}{25}$ and $\frac{2}{25}$, which is $\frac{2}{25}$ seconds.
Q11. For the discrete-time signal $x[n] = u[-n+2] - u[-n-2]$, what is the value of $x[n]$ for $n = 0$?
IntermediateA · 1
At $n=0$: $u[-0+2] = u[2] = 1$ and $u[-0-2] = u[-2] = 0$. So $x[0] = 1 - 0 = 1$.
Q12. The signal $x(t) = r(-t+4)$ is:
IntermediateA · Zero for $t > 4$
The argument $-t+4$ is negative when $t > 4$. Since $r(\text{negative}) = 0$, $x(t) = 0$ for $t > 4$. For $t < 4$, the argument is positive, so $x(t) = -t+4$.
Q13. Which of the following is the correct expression for $x(t) = r(t) \cdot u(-t+3)$ for $t > 3$?
IntermediateC · $x(t) = 0$
For $t > 3$, the argument of $u(-t+3)$ is negative, so $u(-t+3) = 0$. Multiplying by $r(t)$ gives $x(t) = 0$.
Q14. The signal $x(t) = 2u(t) + 3\cos(2\pi t)$ is periodic.
IntermediateB · False
The unit step $u(t)$ is not periodic because it is not repetitive—it is zero for $t0$. Adding a periodic cosine does not make the sum periodic.
Q15. For $x(t) = 3e^{j(4\pi t + \pi/3)} + 4e^{-j(2t + \pi/2)}$, is the signal periodic? If so, what is its fundamental period?
AdvancedB · Periodic with $T_0 = \pi$ seconds
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