Scaling and Shifting

Scaling and Shifting There is a very useful class of shortcuts which allows us to use the known Fourier series of a function f (t) to get the series for a function re­ lated to f (t) by shifts and scale changes. We illustrate this technique with a collection of examples of related functions. We let sq(t) be the standard odd, period 2π square wave. � −1 for − π ≤ t < 0 sq(t) = 1 for 0 ≤ t < π (1) 1 �t π −1 Figure 0: The graph of sq(t), the odd, period 2π square wave. We already know the Fourier series for sq(t). It is � � 4 1 1 4 sin(t) + sin(3t) + sin(5t) + · · · = sq(t) = π 3 5 π 1. sin(nt) (2) n n odd ∑ Shifting and Scaling in the Vertical Direction Example 1. (Shifting) Find the Fourier series of the function f 1 (t) whose graph is shown. 2 �t π Figure 1: f 1 (t) = sq(t) shifted up by 1 unit. Solution. The graph in Figure 1 is simply the graph in Figure 0 shifted upwards one unit. That is, f 1 (t) = 1 + sq(t). Therefore f 1 (t) = 1 + 4 π sin(nt) . n n odd ∑ Example 2. (Scaling) Let f 2 (t) = 2 sq(t). Sketch its graph and find its Fourier series. OCW 18.03SC Solution. 2 π �t −2 Figure 2: Graph of f 2 (t) = 2 sq(t). The Fourier series of f 2 (t) comes from that of sq(t) by multiplying by 2. f 2 (t) = 8 π sin(nt) . n n odd ∑ Example 3. We can combine shifting and scaling along the vertical axis. Let f 3 (t) be the function shown in Figure 3. Write it in terms of sq(t) and find its Fourier series. 1 �t π −1 Figure 3: f 3 (t) = sq(t) shifted by 1 and then scaled by 1/2. 1 1 2 sin nt . Solution. f 3 (t) = (1 + sq(t)) = + ∑ 2 2 π n odd n 2. Scaling and Shifting in t Example 4. (Scaling in time) Find the Fourier series of the function f 4 (t) whose graph is shown. 1 1 �t −1 Figure 4: sq(t) scaled in time. InFigure 4 the point marked 1 on the t-axis corresponds with the point marked π in Figure 0. This shows that f 4 (t) = sq(πt) and therefore we replace t by πt in the Fourier series of sq(t). f 4 (t) = 4 π sin(nπt) . n n odd ∑ OCW 18.03SC Example 5. (Shifting in time) Let f 5 (t) = sq(t + π/2). Graph this function and find its Fourier series. Solution. We have f 5 (t) is sq(t) shifted to the left by π/2. Therefore � � � � 4 cos 3t 4 sin(3t + 3π/2) cos t − sin(t + π/2) + +... = +... f 5 (t) = 3 π 3 π (To simplify the series we used the trig identities sin(θ + π/2) = cos(θ ) and sin(θ + 3π/2) = − cos(θ ) etc.) 1 π/2 �t Figure 5: sq(t) shifted in time. Notice that f 5 (t) is even, and so must have only cosine terms in its series, which is in fact confirmed by the simplified form above.