We’ll begin with a square function, f(t), that has a an amplitude of 1, a start time of 2 seconds and an end time of 4 seconds.
Next, a time shift is demonstrated. Here our function is changed from f(t) to f(t-2). Notice that subtracting 2 from t in the function results in a positive shift of the graph.
In the next two graphs t will be scaled. Scaling t is not quite as intuitive as we may have expected. When we multiply t by 2, corresponding points of the function now occur at 1/2 the time they previously had. When we divide t by 2, each corresponding time on the graph occurs at a t that is now multiplied by 2. Notice that each of these factors directly affects the duration of the signal.
Scaling the amplitude has more intuitive results. If we multiply f(t) by 2, the amplitude of 1 is changed to 2. Multiplying f(t) by 1/2 results in an amplitude of 1/2.
Finally, multiplying t by -1 mirrors our function over the y-axis. Each time now occurs at its negative.
Here we will attempt to convert f(t) into 2*f(.5t+3). The graph of f(t) is shown below.
The easiest way to handle this type of problem without error is to manipulate the function one step at a time. First, I have converted f(t) into 2*f(t). Only the peaks are changed here (by a factor of 2).
Next, I convert 2*f(t) into 2*f( t). Notice how the actually expands our graph duration by a factor of 2 (from a 6 sec duration to a 12 sec duration).
Finally, we move from 2*f( t) to 2*f( t + 3). As shown in the discussion above, this is a time shift. Time shifts can be a little confusing because adding results in a negative shift of our graph. Try to think of it as our signal occurring 3 seconds earlier than before, reading from left to right on the graph. The easiest way to do this part is shift each x-intercept by 3 seconds (to the left, of course).