Signals Fundamentals: 17 flashcards
All 17 flashcards from this kit, free and with no sign-up.
1. What is the fundamental period of x(t) = cos(6πt)?
Show the answer
The fundamental angular frequency is ω0 = 6π rad/s. Period T0 = 2π/ω0 = 2π/(6π) = 1/3 seconds.
2. Is x(t) = e^(j8πt) periodic? If yes, what is its fundamental period?
Show the answer
Yes, it is periodic. ω0 = 8π, so T0 = 2π/(8π) = 1/4 seconds.
3. Determine if x(t) = 2 + sin(4πt) is periodic and find its fundamental period.
Show the answer
Yes, it is periodic. The constant 2 does not affect periodicity. For sin(4πt), ω0 = 4π, so T0 = 2π/(4π) = 1/2 seconds.
4. Is x(t) = 2u(t) + 3cos(2πt) periodic? Explain.
Show the answer
No, it is not periodic. The unit step u(t) is not periodic (it is zero for t<0 and one for t>0), so the sum cannot be periodic.
5. Find the fundamental period of x(t) = 2cos(50πt) + 3sin(25πt).
Show the answer
For cos(50πt): T1 = 2π/(50π) = 1/25. For sin(25πt): T2 = 2π/(25π) = 2/25. Ratio T1/T2 = (1/25)/(2/25) = 1/2, rational, so the sum is periodic. Fundamental period is the LCM of T1 and T2 = 2/25 seconds.
6. Determine if x(t) = 3e^(j[4πt + π/3]) + 4e^(-j[2t + π/2]) is periodic.
Show the answer
First term has ω1 = 4π, T1 = 1/2. Second term has ω2 = 2, T2 = π. Ratio T1/T2 = (1/2)/π = 1/(2π), which is irrational, so the sum is not periodic.
7. Sketch x(t) = u(-t + 4). What is the value of x(t) for t < 4 and t > 4?
Show the answer
u(-t+4) = u(-(t-4)). This is a time-reversed and shifted unit step. It is 1 for -t+4 > 0 i.e., t < 4, and 0 for t > 4. At t=4, it is u(0) = 1 (assuming standard definition).
8. Sketch x(t) = r(-t + 4). What is the expression for x(t) in piecewise form?
Show the answer
r(-t+4) = r(-(t-4)). It is a ramp that decreases as t increases. For -t+4 > 0 (t<4), r(-t+4) = -t+4. For t ≥ 4, it is 0. So x(t) = max(0, -t+4).
9. Sketch x(t) = r(t) - 2r(t-2) + r(t-4). What is the slope of x(t) in the intervals (0,2), (2,4), and (4,∞)?
Show the answer
For t<0: 0. For 0<t<2: slope = 1. For 2<t<4: slope = 1 - 2 = -1. For t>4: slope = 1 - 2 + 1 = 0. So it is a triangular pulse from t=0 to 4 with peak at t=2 of value 2.
10. Sketch x(t) = Π((t-2)/4) + Π(t-2). What is the support (interval where nonzero) of each rectangular pulse?
Show the answer
Π((t-2)/4) is a rectangle of width 4 centered at t=2, so it is 1 for 0<t<4. Π(t-2) is a rectangle of width 1 centered at t=2, so it is 1 for 1.5<t<2.5. The sum is 2 on (1.5,2.5) and 1 on (0,1.5)∪(2.5,4).
11. Sketch x(t) = r(t)·u(-t+3). What is the expression for x(t) for t<3 and t>3?
Show the answer
u(-t+3) is 1 for t<3 and 0 for t>3. So x(t) = r(t) for t<3, and 0 for t>3. At t=3, u(0)=1, so x(3)=3. So x(t) = t for 0≤t<3, and 0 otherwise.
12. Sketch x[n] = u[-n+2] - u[-n-2]. For which integer values of n is x[n] = 1?
Show the answer
u[-n+2] is 1 for -n+2 ≥ 0 ⇒ n ≤ 2. u[-n-2] is 1 for -n-2 ≥ 0 ⇒ n ≤ -2. So x[n] = 1 for n ≤ 2 minus n ≤ -2, which gives n = -1, 0, 1, 2. For n ≤ -3, both are 1, so difference is 0. For n > 2, both 0.
13. For the signal x(t) shown as a triangular pulse with peak amplitude 2 at t=0 and symmetric in [-1,1], sketch x(2t+2). What is the transformation in terms of time shift and scaling?
Show the answer
x(2t+2) = x(2(t+1)). This is a time compression by factor 2 and a left shift by 1. The original support [-1,1] becomes [-1.5, -0.5] (solve -1 ≤ 2t+2 ≤ 1 ⇒ -1.5 ≤ t ≤ -0.5). The peak amplitude remains 2 at t=-1.
14. For the same triangular pulse x(t), sketch x((1/2)t - 2). What is the new support and peak location?
Show the answer
x((1/2)t - 2) = x((1/2)(t-4)). This is a time expansion by factor 2 and a right shift by 4. Original support [-1,1] becomes [2,6] (solve -1 ≤ (1/2)t - 2 ≤ 1 ⇒ 2 ≤ t ≤ 6). Peak amplitude 2 at t=4.
15. For the same triangular pulse x(t), sketch x(-t-2). What is the transformation?
Show the answer
x(-t-2) = x(-(t+2)). This is a time reversal and a left shift by 2. Original support [-1,1] becomes [-3,-1] (solve -1 ≤ -t-2 ≤ 1 ⇒ -3 ≤ t ≤ -1). Peak at t=-2.
16. For the same triangular pulse x(t), sketch 3x(5t). What is the amplitude and support?
Show the answer
3x(5t) is a time compression by factor 5 and amplitude scaling by 3. Original support [-1,1] becomes [-0.2, 0.2] (solve -1 ≤ 5t ≤ 1). Peak amplitude becomes 6 at t=0.
17. Express the signal x(t) = r(t) - 2r(t-2) + r(t-4) in piecewise form.
Show the answer
For t<0: 0. For 0≤t<2: t. For 2≤t<4: t - 2(t-2) = -t + 4. For t≥4: t - 2(t-2) + (t-4) = 0. So x(t) = t for 0≤t<2, -t+4 for 2≤t<4, and 0 otherwise.
Your notes can be a kit like these
Every kit here was made from real study material. Point Veda at a chapter, a PDF, a photo or a link and get notes, flashcards, practice MCQs, games and a scored test — free, in 42 languages.
- Chapter notes
- Flashcards
- Practice MCQs
- Scored tests
- Exam prep
Get Veda free See what students get
Free to start · No ads · Works in 42 languages