#13 Signals & Systems #13 | DT Signals Review & Even/Odd Decomposition

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. Basic sequences

   Corrected solution notes. The audio is the original recording and may contain errors or incomplete phrases. Where they differ, use the corrected written solution.

   u[n] is one for n≥0; δ[n] is one only at zero; r[n]=n u[n].

   The backward difference of the step is δ[n]. The running sum of the step through n−1 is r[n].

   Narration transcript

   Let's begin with a quick review of the fundamental discrete-time signals. The unit step u of n equals one for n greater than or equal to zero and zero for negative n. The unit impulse delta of n is one only at n equals zero. The unit ramp r of n equals n times u of n — it's zero for negative n, and grows linearly for non-negative n. These signals are related: u of n is the running sum of delta of n minus k, and delta of n equals u of n minus u of n minus one. These relationships form a chain: impulse, step, then ramp.

2. 2. Exponential and sinusoid

   The source example K exp(n)u[n] is a growing causal exponential. More generally K aⁿ u[n] grows or decays according to the magnitude of a.

   A sinusoid cos(Ωn+φ) is periodic when Ω/(2π) is rational. With Ω=2π/N, positive integer N is a period; degenerate signals may have a smaller period.

   Narration transcript

   Now two more signal types in discrete time. The exponential signal is f of n equals K times e to the n, times u of n. Here K is the initial amplitude at n equals zero. Unlike continuous time, this signal is only defined at integer values of n, shown as vertical stems. The discrete-time sinusoidal signal is x of n equals K times cosine of two pi over N times n plus Omega zero. The integer N is the period — the number of samples in one full cycle. The angular frequency is Omega equals two pi over N radians per sample. Just like in continuous time, K controls amplitude and Omega zero controls the phase shift.

3. 3. Step compositions

   u[n−2] starts at n=2; u[n−2]+u[n−4] increases from one to two at n=4.

   u[n−2]−2u[n−6] is one on 2≤n≤5 and −1 from n=6 onward.

   r[n]−r[n−3] increases to three, then stays at three.

   Narration transcript

   Let's practice writing and plotting discrete-time signal compositions. In the first panel, u of n minus two is simply the unit step shifted right by two — it equals one starting from n equals two. In the second panel, u of n minus two plus u of n minus four creates a staircase: it's zero before n equals two, one from n equals two to three, and two from n equals four onward. The third panel shows u of n minus two minus two u of n minus six — it's one from n equals two to five, then drops to negative one from n equals six onward. The fourth panel shows r of n minus r of n minus three: the ramp rises from zero to three, then stays flat at three — the second ramp cancels the growth.

4. 4. Ramp composition

   For f[n]=r[n]+r[n−2]−2r[n−4], values are zero for n<0; n for 0≤n≤2; 2n−2 for 2≤n≤4; and six for n≥4.

   The formulas agree at the shared endpoints.

   Narration transcript

   Here's a more complex example: f of n equals r of n plus r of n minus two minus two r of n minus four. Let's build it step by step. First, r of n gives us a ramp with slope one starting at the origin. Second, r of n minus two adds another ramp starting at n equals two with slope one. When we combine them, the slope doubles to two from n equals two onward. Finally, minus two r of n minus four subtracts a ramp with slope two starting at n equals four, which exactly cancels the total slope. The result: slope one from zero to two, slope two from two to four, then flat at six for n greater than or equal to four. This is the key technique: adding and subtracting shifted ramps controls both the slopes and the breakpoints.

5. 5. Correct special cases

   $\displaystyle \delta \left[2n\right]=\delta \left[n\right].$

   For every integer n, 2n−3 is an integer unequal to zero; therefore δ[2n−3] is the well-defined all-zero sequence.

   r[n²]=n² for all integer n. This is not the same operation as squaring r[n] at negative indices.

   Narration transcript

   Two interesting special cases from discrete-time signal manipulation. First, consider delta of two n. Since the impulse delta of n is nonzero only at n equals zero, substituting two n for n still gives a nonzero value only when two n equals zero — that is, n equals zero. So delta of two n equals delta of n. However, delta of two n minus three would require n equals three halves, which is not an integer. So this expression is always zero — it's undefined in discrete time. Second, consider r of n squared. Since n squared is always non-negative for any integer n, the unit step embedded in the ramp is always active. So r of n squared equals n squared for all integer n — it traces out a parabola in both directions.

6. 6. Even and odd parts

   xe[n]=(x[n]+x[−n])/2 and xo[n]=(x[n]−x[−n])/2.

   The first is even, the second odd, and their sum reconstructs x[n]. The odd part is zero at the origin.

   Narration transcript

   Any signal can be decomposed into even and odd parts. The even part is x sub e of t equals one half times x of t plus x of negative t. The odd part is x sub o of t equals one half times x of t minus x of negative t. The even part satisfies x sub e of negative t equals x sub e of t — it's symmetric about the origin. The odd part satisfies x sub o of negative t equals negative x sub o of t — it's antisymmetric, and always zero at the origin. Together they reconstruct the original: x of t equals x sub e of t plus x sub o of t. Let's see an example. We start with an asymmetric signal x of n. We mirror it to get x of negative n. Averaging gives the symmetric even part. Subtracting and halving gives the antisymmetric odd part. Every signal decomposes this way.

7. 7. Review

   

   Check the integer argument before evaluating a sequence. No half-integer impulse location is needed to define δ[2n−3].

   Use shifted steps and ramps for piecewise construction, then verify symmetry and reconstruction.

   Narration transcript

   Today we reviewed the fundamental discrete-time signals: step, impulse, and ramp, along with their DT exponential and sinusoidal counterparts. We practiced composing signals from shifted steps and ramps, and saw special cases like delta of two n equals delta of n, and r of n squared equals n squared. Finally, we learned that any signal splits into even and odd parts using the formulas x sub e equals one half of x plus x of negative t, and x sub o equals one half of x minus x of negative t. In the next lesson, we'll move to discrete-time convolution — the operation that defines how LTI systems process signals.

Source video: Signals & Systems #13 | DT Signals Review & Even/Odd Decomposition (6:28)