Hello. In this video, we review the structure of sinusoids and the parameters of sine and cosine functions. A typical sinusoid would take one of two forms: a cosine function or a sine function. The input argument will be in this form: Ωt + φ, and you can see here that the sinusoid depends on the time. Here we have a scaling factor A. We call the scaling factor the amplitude. The units of the amplitude depend on the nature of the signal. So if the sinusoid here represents a voltage signal, then it would be in volts. If the sinusoid describes a current, it should be in amps, and so on and so forth.

Ω is the angular frequency, and it has units of rad/s. φ is the phase shift. It has units of radians or degrees, which actually makes sense because the input argument should have those units. Sinusoids are trigonometric functions, and the input argument should be a phase. The units for the time are seconds, so if you'd like to have a phase, the units of Ω should be rad/s so that when you multiply (rad/s × s), you get units of an angle or a phase. The same applies to φ, which is the phase shift—it has units of an angle.

In this plot here, we have the two functions: sin(t) on the left and cos(t) on the right. Both have the same frequency and the same magnitude (amplitude). Here, the amplitude is one, which makes sense because when we studied sinusoids, we know the relationship between the sine and cosine functions. We know that cos(θ) = sin(θ + 90°). So, for both sine and cosine at the same frequency, you would have the same shape of oscillation but with a shift of 90 degrees. The sine here, with no phase shift, starts at the origin with a magnitude of zero, whereas the cosine starts at the origin with a magnitude of one.

Now, if I have two sinusoids oscillating at the same frequency but shifted in time like that, this time shift corresponds to a phase shift. Here, the two figures show sinusoids that oscillate at the same frequency with the same amplitude, but since there is a time difference between the two signals, this time difference actually corresponds to a phase shift.

Now, let's formally introduce those parameters—magnitude, frequency, and phase shift—and study their effects on our sinusoid. The amplitude of a sinusoid determines the distance from the average value of our sinusoid to its peak. For the signal, this point here would give me the peak. Here, we have a magnitude of one. In the right figure, the average value is zero, so the distance from the average value to the peak gives us the magnitude. Here, we have a magnitude of two. So, basically, the magnitude is the scale of our sinusoid.

Next, we have another parameter: the period, which is tightly related to the frequency. By definition, the period (symbol T) is defined as the time it takes our sinusoid to complete one cycle. The sinusoid repeats, so the time to complete one cycle is defined as the period. The inverse of this period defines the frequency. The frequency, by definition, is the number of cycles per second, and its unit is Hertz (Hz). So, if you have 1 Hz, this corresponds to one cycle per second.

Here, we have a couple of examples. Within a time frame, if I have one cycle, then the frequency is 1 Hz. In the second example, I have two cycles (cycle number one and cycle number two), so we have two cycles per second, meaning the frequency here is 2 Hz. In the last example, we have 60 repetitions (or 60 cycles) in one second, so the frequency is 60 Hz.

By definition, the frequency is the inverse of the period, so f (in Hz) = 1/T (in seconds). The angular frequency (Ω) is related to the frequency. The frequency tells us how many cycles occur per second, while the angular frequency can be thought of as the angular velocity of your sinusoid. It has units of rad/s and is defined as Ω = 2πf. We can also replace f with 1/T, so Ω = 2π/T, again with units of rad/s.

Now, what is the effect of changing Ω? Since Ω is linearly proportional to f, it has the same effect: increasing Ω increases the oscillation rate of our signal. The higher the frequency, the faster the oscillations; the lower the frequency, the slower the oscillations.

Now, for the parameter φ (the phase shift), the phase shift represents a time shift of our sinusoid. Depending on the sign of φ, you determine whether your signal shifts to the left or the right. In the example below, I have a signal with a positive phase shift. You can see that with a positive phase shift, our sine is shifted to the left, meaning it starts before zero. On the other hand, if I have a negative phase shift, the sine starts at a later time than zero, so we say it is lagging. A sine should start at zero, but now we have a time shift, so the phase shift encodes the time shift information.

What is the actual time shift corresponding to a phase shift of φ? We have a relationship between the time shift and the phase shift: the time shift (t₀) = φ/Ω. We know that Ω = 2πf, so t₀ = φ/(2πf). Replacing f with 1/T, we also have t₀ = (φT)/(2π). These three expressions tell you how much time you're going to shift your sinusoid given a phase shift of φ.

One last definition for sinusoids is the instantaneous value, which is the value of the sinusoid function at a particular time instant. For example, in the figure, at t = 0.5 ms, the instantaneous value is 20 V, and at t = 2 ms, it is 35 V. The instantaneous value is simply the value of our sinusoid at a specific time. To obtain it, if I ask you for the instantaneous value at time t₀, you simply plug t₀ into the expression for the sinusoid. The expression is A·cos(Ωt + φ). To get the instantaneous value at t₀, plug t = t₀ into the function and solve.

Let's wrap up with a couple of examples to reinforce these definitions. On the right, we have a signal modeled as A·sin(Ωt + φ). We would like to determine the parameters of the cosine: amplitude, angular frequency, and phase shift. We can start with the frequency, but it's easier to first find the period. If you consider this point, the sinusoid completes one cycle here, so the time it takes is the period (T). From the figure, T = 20 ms. From this, we can obtain f = 1/T = 1/(20 ms) = 50 Hz. Once we have f, we can find Ω = 2πf = 100π rad/s. Now we have Ω.

The magnitude is also straightforward. Here, it is the deviation of the sinusoid from its average value. The sinusoid is symmetric around the origin, so the distance from the average value to the maximum is the magnitude (A = 10 V). What remains is the phase shift (φ). By inspection, the signal is a cosine function because the peak starts at zero, and we know that cos(θ) = sin(θ + 90°), so φ should be 90°. If we didn't know this, we could solve for φ using the relationship φ = t₀·Ω, where t₀ is the time shift. Here, the sine function is shifted by t₀ = 5 ms, so φ = (5 ms) × (100π rad/s) = 0.5π = 90°.

Now, let's move to the second example. Here, we have a sine wave with f = 100 Hz and φ = 0. We don't know the magnitude, but we know the instantaneous value is 100 V at t = 1.25 ms. The question asks for the voltage at t = 2.5 ms. The signal is V(t) = A·sin(2π·100·t). Plugging in t = 1.25 ms, we solve for A: A = 100 / sin(2π·100·1.25×10⁻³) = 141.42 V. Now, at t = 2.5 ms, V(t) = 141.42·sin(2π·100·2.5×10⁻³) = 141.42 V (since sin(π/2) = 1).

In conclusion, in this video, we reviewed the basic structure of a sinusoid function. It has three main parameters: amplitude (controls the scale), angular frequency (controls the speed of oscillation), and phase shift (determines the time shift). We also learned about the instantaneous value, which is the value of the sinusoid at a specific time. This concludes our video. Thank you so much!