Signals and Systems · Continuous-time step
#07 Signals & Systems #07 | Unit Step u(t) & Ramp r(t) - CT and DT Definitions
Corrected solution notes. The audio is the original recording and may contain errors or incomplete phrases. Where they differ, use the corrected written solution.
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Corrected solution notes. The audio is the original recording and may contain errors or incomplete phrases. Where they differ, use the corrected written solution. Signals & Systems #07 | Unit Step u(t) & Ramp r(t) - CT and DT Definitions
Written solution and narration transcript(shows the full solution)
Below are all the lines written in the notebook together with the full narration transcript.
1. Continuous-time step
Corrected solution notes. The audio is the original recording and may contain errors or incomplete phrases. Where they differ, use the corrected written solution.This lesson chooses u(t)=0 for t≤0 and u(t)=1 for t>0.The value at the switching instant must be treated consistently in pointwise examples.Narration transcript
The first elementary signal we study is the unit step function u of t. It is defined as a piecewise function: u of t equals one when t is greater than zero, and zero when t is less than or equal to zero. Graphically, it is flat at zero for all negative time, then jumps instantaneously to one at t equals zero and stays there forever. This sudden jump makes it ideal for modeling signals that switch on at a specific moment, like flipping a switch. The value at exactly t equals zero is sometimes debated, but we will use the simpler first definition where u of zero equals zero.
2. Discrete-time step
For the discrete-time convention, u[n]=1 at n≥0 and zero at negative indices.Thus the continuous-time choice u(0)=0 and the discrete-time choice u[0]=1 are different conventions.Narration transcript
Now let us compare the continuous-time and discrete-time versions side by side. For continuous time, u of t is a horizontal line at zero for t less than zero, with an instantaneous jump to one at t equals zero. For discrete time, u of n equals one when n is greater than or equal to zero, and zero for negative n. In the stem plot, we see dots at zero for negative n, and dots at one for n equals zero, one, two, three, and so on. Notice that the DT version has no ambiguity at n equals zero: it simply equals one.
3. Continuous-time ramp
r(t)=t for t≥0 and zero for t<0.Its integral relation with the step is valid; its classical derivative at the corner is not defined. The derivative equals the step away from zero and as a distribution independently of the chosen point value.Narration transcript
The second elementary signal is the unit ramp function r of t. It equals t when t is greater than or equal to zero, and zero otherwise. We can write this compactly as r of t equals t times u of t. Graphically, it is flat at zero for negative time, then rises with slope one starting at the origin. The ramp is the integral of the step: r of t equals the integral of u of tau d tau from negative infinity to t. Conversely, the step is the derivative of the ramp: u of t equals d r of t d t. This derivative-integral relationship will become very important when we study the impulse function.
4. Correct the discrete running sum
With r[n]=n u[n], summing u[k] from negative infinity through n gives (n+1)u[n], not r[n].To obtain r[n], sum through n−1. Equivalently, r[n+1]−r[n]=u[n].Narration transcript
Comparing continuous and discrete ramp functions: for continuous time, r of t is a straight line with slope one for t greater than or equal to zero. For discrete time, r of n equals n times u of n, which gives us the values zero, one, two, three, four at n equals zero, one, two, three, four. The stem plot shows these linearly increasing values. Just like in continuous time, the DT ramp is the running sum of the DT step: r of n equals the sum of u of k from k equals negative infinity to n.
5. Shift the signals
u(t−t0) switches just after t0 under the stated point convention; r(t−t0) starts its ramp at t0.A positive t0 delays, and a negative t0 advances the signal.Narration transcript
We can shift both step and ramp functions in time. A delayed unit step u of t minus t zero switches on at t equals t zero instead of at the origin. For example, u of t minus two turns on at t equals two. An advanced step u of t plus three turns on at t equals negative three. The same logic applies to ramps: r of t minus t zero starts ramping at t equals t zero with slope one. These shifted elementary signals are building blocks for constructing more complex signals.
6. Staircase endpoints
For y(t)=u(t−1)+u(t−2), the value is zero at t≤1, one at 1<t≤2, and two at t>2.In particular y(2)=1 under u(0)=0. Endpoint values must agree with the definition used at the start.Narration transcript
Let us build a composite signal. Consider y of t equals u of t minus one plus u of t minus two. The first component, u of t minus one, is a step that turns on at t equals one. The second component, u of t minus two, turns on at t equals two. When we add them together, we get a staircase: y of t is zero for t less than one, one for t between one and two, and two for t greater than or equal to two. By adding more shifted steps, we can build any staircase pattern. For example, adding u of t minus three gives us a three-level staircase.
7. Review
Steps model switching and ramps model linear growth.Keep continuous-time endpoint conventions separate from discrete-time summation limits; both affect pointwise identities.Narration transcript
Let us recap. The unit step u of t switches from zero to one at the origin, while u of n does the same at integer n equals zero. The unit ramp r of t rises linearly with slope one, and r of n increases by one at each integer step. Both can be shifted in time to create delayed or advanced versions. And by combining shifted steps, we build staircase signals. In the next lesson, we will meet the most important elementary signal: the impulse function delta of t.
Source video: Signals & Systems #07 | Unit Step u(t) & Ramp r(t) - CT and DT Definitions (4:52)