--- title: "Unit step and delta: CT and DT" subject: "Signals and Systems" chapter: 1 key: "semester_4/signals_and_systems:unit-step-delta-relation" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/unit-step-delta-relation site: BCT Engineering --- # Unit step and delta: CT and DT Signals and Systems · Chapter 1: Signals - Asked in: 2072 Magh, 2+3 marks; 2083 Bhadra (new course), 5 marks (asked 2 times) - Video: [10- Signals and Systems: DT Unit Impulse and Unit Step Function](https://www.youtube.com/watch?v=w5dq3PoPJmI) — Prof Sarmad - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/unit-step-delta-relation - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1 ## Question Explain unit step and unit delta signal both in continuous time and discrete time. Also, explain the relationship between the unit step and delta function. ## Short answer CT unit step u(t)=1 for t>0, 0 for t<0; delta δ(t) is zero except at t=0 with unit area. In DT, u[n]=1 for n≥0, 0 otherwise; δ[n]=1 at n=0, 0 elsewhere. Relationship: u(t)=∫_{-∞}^t δ(τ)dτ, δ(t)=du/dt; in DT, u[n]=∑_{k=-∞}^n δ[k], δ[n]=u[n]−u[n−1]. ## Answer #### Continuous time **Unit step** $u(t)$: $$ u(t) = \begin{cases} 1, & t > 0 \\ 0, & t < 0 \end{cases} $$ It is discontinuous at $t = 0$ (value there is undefined or taken as 1/2). It is used to switch signals on at $t = 0$. **Unit impulse (Dirac delta)** $\delta(t)$: zero everywhere except at $t = 0$, with unit area: $$ \delta(t) = 0 \text{ for } t \neq 0, \qquad \int_{-\infty}^{\infty}\delta(t)\,dt = 1 $$ It is the limit of a rectangular pulse of width $\Delta$ and height $1/\Delta$ as $\Delta \to 0$. **Sifting property:** $\int x(t)\delta(t-t_0)\,dt = x(t_0)$. #### Discrete time $$ u[n] = \begin{cases} 1, & n \ge 0 \\ 0, & n < 0 \end{cases} \qquad \delta[n] = \begin{cases} 1, & n = 0 \\ 0, & n \neq 0 \end{cases} $$ $\delta[n]$ is a simple sequence with value exactly 1 at $n = 0$ (no limiting process needed). Any sequence can be written as $x[n] = \sum_k x[k]\delta[n-k]$. ```text u(t) delta(t) 1 +--------- ^ area = 1 | | --+--------> t --+--------> t 0 0 u[n] 1 o o o o ... | | | | -----o--o--+--+--+--+---> n -2 -1 0 1 2 3 delta[n] 1 o | --o--+--o--> n -1 0 1 ``` #### Relationship **Continuous time:** $$ u(t) = \int_{-\infty}^{t}\delta(\tau)\,d\tau, \qquad \delta(t) = \frac{du(t)}{dt} $$ The step is the running integral of the impulse; the impulse is the derivative of the step (the jump of height 1 at $t=0$). **Discrete time:** $$ u[n] = \sum_{k=-\infty}^{n}\delta[k] = \sum_{k=0}^{\infty}\delta[n-k], \qquad \delta[n] = u[n] - u[n-1] $$ The step is the running sum of the impulse; the impulse is the first difference of the step. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which describes the CT unit impulse delta(t)? A. Value 1 at t = 0 only B. Zero for t != 0 with unit area C. Value 1 for t > 0 D. Unit height pulse of width 1 Answer: B. delta(t) is zero away from the origin and its integral over all t equals 1. (Key point: CT impulse definition) 2. How does delta[n] differ from delta(t)? A. It is exactly 1 at n = 0 and 0 elsewhere, with no limit process B. It has infinite height C. It is defined only for n > 0 D. It has area 1 in the continuous sense Answer: A. delta[n] is an ordinary sequence with value 1 at n = 0. (Key point: DT impulse) 3. The relationship between u(t) and delta(t) is: A. u(t) = d delta/dt B. delta(t) = du/dt and u(t) = integral of delta C. delta(t) = integral of u D. They are unrelated Answer: B. The step is the running integral of the impulse; the impulse is the derivative of the step. (Key point: CT relationship) 4. In DT, delta[n] equals: A. u[n] + u[n-1] B. u[n] - u[n-1] C. u[n] u[n-1] D. sum of u[k] Answer: B. The first difference of the step is 1 only at n = 0, which is delta[n]. (Key point: DT relationship) 5. The sifting property states integral of x(t) delta(t - t0) dt equals: A. x(0) B. x(t0) C. 1 D. t0 Answer: B. The impulse picks out the value of x at the instant t0. 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