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Chapter 2

Complex Numbers and Exponentials

Abstract

This chapter introduces the importance of complex numbers and exponentials which are an integral part of digital communications and digital signal processing (DSP). The reason of delving deep into this area leads the readers to two-dimensional number plane, which help to understand DSP. The operators such as addition, subtraction, and multiplication take people through the complex side of numbers which are explained in two dimensions. The use of complex conjugate and complex exponential—in the number plane—allows readers understand the relationship of basic characteristics of numbers and its treatment. The chapter further looks into measuring angles in radians, which leads to the use of pi and the equivalence of the degrees and radians in measuring angles. DSP therefore uses this concept in later stages to understand the clockwise and anticlockwise movement around the circle.

Keywords

Complex conjugate; Complex exponential; Digital signal processing; Euler equation; Polar representation

Complex numbers are one of those things many of us were taught a long time ago and have long since forgotten. Unfortunately, they are important in digital communications and digital signal processing (DSP), so we need to resurrect them.

What we were taught and some of us vaguely remember is that a complex number has a “real” and “imaginary” part, and the imaginary part is the square root of a negative number, which is really a nonexistent number. This right away sounds fishy, and while it's technically true, there is a much more intuitive way of looking at it.

The whole reason for “complex numbers” is that we are going to need a two dimensional number plane to understand DSP. The traditional number line extends from plus infinity to minus infinity, along a single line. To represent many of the concepts in DSP, we need two dimensions. This requires two orthogonal axes, like a North–South line and an East–West line. For the arithmetic to work out, one line, usually depicted as the horizontal line, is the real number line. The other vertical line is the imaginary line. All imaginary numbers are prefaced by “j”, which is defined as the square root of −1. Do not get confused by this imaginary number stuff, but rather view “j” as an arbitrary construct we will use to differentiate the horizontal axis (normal) numbers from those on the vertical axis. This is the essence of this whole chapter.

As depicted in Fig. 2.1, any complex number Z has a real and imaginary part, and is expressed as X + j·Y, or just X + jY. The value of X and Y for any point is determined by the distance one must travel in the direction of each axis to arrive at the point. It can also be visualized as a point on the complex number plane, or as a vector originating at the origin and terminating at the point. We need to be able to do arithmetic with complex numbers. There has to be a way to keep track the vertical and horizontal components. That is where the “j” comes in (in some texts, “i” is used instead).

2.1. Complex Addition and Subtraction

Adding and subtracting is simple—just add and subtract the vertical and horizontal components separately. The “j” helps us keep from mixing the vertical and horizontal components. For example:

$(3 + j 4) - (1 - j 6) = 2 + j 10$ $(3 + j 4) - (1 - j 6) = 2 + j 10$

Scaling a complex number is just simply scaling each component:

$4 \cdot (3 + j 4) = 12 + j 16$ $4 \cdot (3 + j 4) = 12 + j 16$

2.2. Complex Multiplication

Multiplication gets a little trickier and is harder to visualize graphically. Here is the way the mechanics of it work:

$(A + jB) \cdot (C + jD) = A \cdot C + jB \cdot C + A \cdot jD + jB \cdot jD = AC + jBC + jAD + j^{2} BD$ $(A + jB) \cdot (C + jD) = A \cdot C + jB \cdot C + A \cdot jD + jB \cdot jD = AC + jBC + jAD + j^{2} BD$

Now remember that j² is by definition equal to −1. After collecting terms:

$AC + jBC + jAD + j^{2} BD = AC + jBC + jAD - BD = (AC - BD) + j (BC + AD)$ $AC + jBC + jAD + j^{2} BD = AC + jBC + jAD - BD = (AC - BD) + j (BC + AD)$

The result is another complex number, with AC−BD being the real part, and BC + AD being the imaginary part (remember, imaginary just means the vertical axis, while real is the horizontal axis). This result is just another point on the complex plane.

This mechanics of this arithmetic may be simple, but we need to be able to visualize what is really happening. To do this, we need to introduce polar (R-Ω) representation. Up until now, we have been using Cartesian (X-Y) coordinates, which means each location on the complex number plane is specified by the distance along each of the two axes (like longitude and latitude on the earth's surface). What polar representation does is to replace these two parameters, which can specify any point on the complex plane, with another set of two parameters, which also can specify any point on the complex plane. The two new parameters are the magnitude and angle. The magnitude is simply the length of the line or vector from the origin to the point. The angle is defined as the angle of this line starting at the positive X axis and arcing counterclockwise.

Figure 2.2 Magnitude-angle complex number plane.

This is shown in Fig. 2.2, where the same point Z = X + jY is identified by having radius R (length of vector from origin to the point) with angle Ω specified a counterclockwise from positive real axis.

Any point Z on the graph may be specified as X + jY or R with angle Ω.

The relationships between these go back to basic high school math. Consider the right triangle formed in Fig. 2.3, with sides X, Y, R, and angle Ω.

Remember the Pythagorean theorem? For any right triangle, X² + Y² = R².

Also, remember that sine is defined as length of opposite side divided by hypotenuse, and cosine is defined as length of adjacent side divided by hypotenuse (but must be a right triangle). Tangent is defined as opposite over adjacent side. So we get the following relationships, which can be used to convert between Cartesian (X, Y) and polar (R, Ω):

Figure 2.3 Cartisian and polar conversion relationships.

2.3. Polar Representation

$\sin (Ω) = (Y / R) \Rightarrow Y = R \cdot \sin (Ω)$ $\sin (Ω) = (Y / R) \Rightarrow Y = R \cdot \sin (Ω)$

$\cos (Ω) = (X / R) \Rightarrow X = R \cdot \cos (Ω)$ $\cos (Ω) = (X / R) \Rightarrow X = R \cdot \cos (Ω)$

$X^{2} + Y^{2} = R^{2} \Rightarrow R = sqrt (X^{2} + Y^{2})$ $X^{2} + Y^{2} = R^{2} \Rightarrow R = sqrt (X^{2} + Y^{2})$

$\tan (Ω) = (Y / X) \Rightarrow Ω = arctan (Y / X)$ $\tan (Ω) = (Y / X) \Rightarrow Ω = arctan (Y / X)$

The reason for this little foray into polar representation is that multiplication (and division) of complex numbers is very easy in polar form, and that angles in the complex plane can be easily visualized in polar form.

2.4. Complex Multiplication Using Polar Representation

We will define two points, Z₁ and Z₂.

$Z_{1} = R_{1} angle (Ω_{1}) Z_{2} = R_{2} angle (Ω_{2})$ $Z_{1} = R_{1} angle (Ω_{1}) Z_{2} = R_{2} angle (Ω_{2})$

$Z_{1} \cdot Z_{2} = (R_{1} \cdot R_{2}) angle (Ω_{1} + Ω_{2})$ $Z_{1} \cdot Z_{2} = (R_{1} \cdot R_{2}) angle (Ω_{1} + Ω_{2})$

What this means is that with any two complex numbers, the magnitude, or distance from the origin to the radius, gets multiplied together to form the new magnitude. This makes sense intuitively. The angles of the two complex numbers get added together to form the new angle. Not so intuitive, so let us try a few examples to get the hang of it.

Let us use real numbers to start–a real number is just a complex number with the “j” part equal to zero. Real numbers are “simplified” version of complex numbers, so any arithmetic rules on a complex number had better work with the real numbers.

$Z = R angle (Ω) - for real number, the angle must be equal to 0$ $Z = R angle (Ω) - for real number, the angle must be equal to 0$

Then X = R·cos(0) or X = R and Y = R·sin(0) or Y = 0. This is what we expect–the Y portion must be zero for the number to lie on the X (real number) axis.

Now consider two complex numbers, both with angle zero.

$Z_{1} = R_{1} angle (0) {, Z}_{2} = R_{2} angle (0) with the product$ $Z_{1} = R_{1} angle (0) {, Z}_{2} = R_{2} angle (0) with the product$

$Z_{1} \cdot Z_{2} = R_{1} \cdot R_{2} angle (0 + 0) = R_{1} \cdot R_{2} (reverts to traditional multiplication)$ $Z_{1} \cdot Z_{2} = R_{1} \cdot R_{2} angle (0 + 0) = R_{1} \cdot R_{2} (reverts to traditional multiplication)$

Now consider another set complex numbers, both with angles of 180°degrees.

$Z_{1} = R_{1} angle (180) {, Z}_{2} = R_{2} angle (180)$ $Z_{1} = R_{1} angle (180) {, Z}_{2} = R_{2} angle (180)$

From relation above, Y = R·sin (Ω), X = R·cos (Ω). With angle of 180°degrees, Y = 0, X = −R, meaning the point Z lies on the negative part of real axis. Z is simply a negative real number. If we multiply two real negative numbers, we know that we should get a real positive number. Let us check using complex multiplication.

$Z_{1} \cdot Z_{2} = R_{1} \cdot R_{2} angle (180 + 180) = R_{1} \cdot R_{2} angle (360) .$ $Z_{1} \cdot Z_{2} = R_{1} \cdot R_{2} angle (180 + 180) = R_{1} \cdot R_{2} angle (360) .$

The angle 360°degrees is all the way around the circle and equal to 0°degree.

$Z_{1} \cdot Z_{2} = R_{1} \cdot R_{2} angle (360) = R_{1} \cdot R_{2} angle (0) = R_{1} \cdot R_{2} = X_{1} \cdot X_{2}$ $Z_{1} \cdot Z_{2} = R_{1} \cdot R_{2} angle (360) = R_{1} \cdot R_{2} angle (0) = R_{1} \cdot R_{2} = X_{1} \cdot X_{2}$

The result is a real positive number.

There is another point to all these exercises, which is to explain why we chose something strange like j equals square root of −1 to designate the vertical axis in the complex number plane. Be a little more patient—we are almost there.

There are 4 “special” angles—0, 90, 180, and 270 degrees. Notice that:

Z = R angle (0) = X degrees is a positive real number on positive real axis

Z = R angle (90) = Y degrees is a positive imaginary number on positive imaginary axis

Z = R angle (180) = −X degrees is a negative real number on negative real axis

Z = R angle (270) = −Y degrees is a negative imaginary number on negative imaginary axis

If we add 360 degrees to any complex number, it wraps all the way around the circle. Or we can have a negative angle, which means just going backwards (clockwise) around the circle.

Z = R angle (0) = R angle (360) = R angle (720)…

Z = R angle (120) = R angle (480) = R angle (840)…

Z = R angle (90) = R angle (−270) = R angle (−630)…

Z = R angle (−53) = R angle (307) = R angle (667)…

We now know when multiplying two complex numbers, the magnitudes R are multiplied and the angles Ω are summed. Now let us consider a few example cases to illustrate how this imaginary “j” operator helps us.

Imagine two complex numbers with only an imaginary component. They are both located on the positive imaginary axis.

$Z_{1} = {jY}_{1} {, Z}_{2} = {jY}_{2} ({real parts X}_{1} = X_{2} = 0)$ $Z_{1} = {jY}_{1} {, Z}_{2} = {jY}_{2} ({real parts X}_{1} = X_{2} = 0)$

$Z_{1} \cdot Z_{2} = {jY}_{1} \cdot {jY}_{2} = j \cdot j \cdot Y_{1} \cdot Y_{2}$ $Z_{1} \cdot Z_{2} = {jY}_{1} \cdot {jY}_{2} = j \cdot j \cdot Y_{1} \cdot Y_{2}$

Recall we defined j = sqrt(−1), so j² = −1

$Z_{1} \cdot Z_{2} = - (Y_{1} \cdot Y_{2}), a negative real number$ $Z_{1} \cdot Z_{2} = - (Y_{1} \cdot Y_{2}), a negative real number$

Or equivalently,

$Z_{1} = R_{1} angle (90) {, Z}_{2} = R_{2} angle (90)$ $Z_{1} = R_{1} angle (90) {, Z}_{2} = R_{2} angle (90)$

$Z_{1} \cdot Z_{2} = R_{1} \cdot R_{1} angle (180) = - (R_{1} \cdot R_{2}) = - (Y_{1} \cdot Y_{2}) {since X}_{1} = X_{2} = 0$ $Z_{1} \cdot Z_{2} = R_{1} \cdot R_{1} angle (180) = - (R_{1} \cdot R_{2}) = - (Y_{1} \cdot Y_{2}) {since X}_{1} = X_{2} = 0$

You can experiment with other combinations, but what you will find is that the arithmetic of adding angles around the circle when multiplying complex numbers works out perfectly when we designate the positive imaginary axis with j, and the negative imaginary axis with–j.

By visualizing this business of going round the circle, you can see by inspection that:

Multiply two positive real numbers, both angles = 0, result has angle of zero.

$3 \cdot 5 = 15$ $3 \cdot 5 = 15$

Multiply two negative real numbers, both angles = 180 (or −180), result has angle of zero (or 360).

$- 3 \cdot - 5 = 15$ $- 3 \cdot - 5 = 15$

Multiply a positive real number (angle 0) with a negative real number (angle 180), result has angle of 180—a negative real number.

$3 \cdot - 5 = - 15$ $3 \cdot - 5 = - 15$

Multiply a positive real number (angle 0) with a positive imaginary number (angle 90), result has angle of 90—an imaginary number.

$j 3 \cdot 5 = j 15$ $j 3 \cdot 5 = j 15$

Multiply a positive imaginary number (angle 90) with a positive imaginary number (angle 90), result has angle of 180—a negative real number.

$j 3 \cdot j 5 = j^{2} \cdot 15 = - 15$ $j 3 \cdot j 5 = j^{2} \cdot 15 = - 15$

Multiply a negative imaginary number (angle −90) with a negative imaginary number (angle −90), result has angle of 180—a negative real number.

$- j 3 \cdot - j 5 = (- j) \cdot (- j) \cdot 15 = - (- (- (15))) = - 15$ $- j 3 \cdot - j 5 = (- j) \cdot (- j) \cdot 15 = - (- (- (15))) = - 15$

Multiply a positive imaginary number (angle 90) with a negative imaginary number (angle −90), result has angle of 0—a positive real number.

$j 3 \cdot - j 5 = j \cdot (- j) \cdot 15 = - (- (15)) = 15$ $j 3 \cdot - j 5 = j \cdot (- j) \cdot 15 = - (- (15)) = 15$

2.5. Complex Conjugate

The last example illustrates a special case. Every number Z = R angle (Ω) has, what is called, a complex conjugate, Z∗ = R angle (−Ω). In the example above, Ω = 90, but Ω can be any angle. The “∗” symbol is the complex conjugate symbol and means to take the point Z and mirror it across the X axis as shown in Fig. 2.4.

So Z = X + jY has conjugate Z∗ = X–jY. We just negate or reverse sign of imaginary part of a number to get its conjugate, or if in polar form, just negate the sign of the angle.

$Z = R angle (Ω), Z^{*} = R angle (Ω)$ $Z = R angle (Ω), Z^{*} = R angle (Ω)$

A special property of the complex conjugate is that for any complex number:

$Z \cdot Z^{*} = R angle (Ω) \cdot R angle (- Ω) = R^{2} angle (0) .$ $Z \cdot Z^{*} = R angle (Ω) \cdot R angle (- Ω) = R^{2} angle (0) .$

In other words, when you multiply a number by its conjugate, the product is a real number, equal to the magnitude squared. This will become important in digital communication, because it can be used to compute the power of a complex signal.

To summarize, we have tried to show that the imaginary numbers which are used to form things called complex numbers are really not so complex, and imaginary is really a very misleading description. What we have really been after is to create a two dimensional number plane, and define a set of expanded arithmetic rules to manipulate the numbers in it. Now we are ready to move onto the next topic, the complex exponential.

2.6. The Complex Exponential

The complex exponential has an intimidating sound to it, but in reality, it is very simple to visualize. It is simply the unit circle (radius = 1) on the complex number plane (Fig. 2.5).

Any point on the unit circle can be represented by “e” or raised to the power (j·angle) or more also expressed e^jΩ, which is called a complex exponential function. A few examples should help.

Let the angle Ω = 0°degree. Anything raised to the power 0 is equal to 1. This checks out, since this is the Point 1 on the positive real axis.

Let angle Ω = 90°degrees. The complex exponential is e^j90. This is the point j on the positive imaginary axis. We need a way to evaluate the complex exponential to show this. This leads to the Euler equation. This equation can easily be derived using series Taylor expansion for exponential, but we have promised to minimize the math. But the result is:

$e^{j Ω} = \cos (Ω) + j \sin (Ω)$ $e^{j Ω} = \cos (Ω) + j \sin (Ω)$

Let us try exp(j90) again. Using Euler equation

$e^{j 90} = \cos (90) + j \sin (90) = 0 + j \cdot 1 = j$ $e^{j 90} = \cos (90) + j \sin (90) = 0 + j \cdot 1 = j$

Imagine the point Z = e^jΩ with the angle Ω starting at 0°degree and gradually increasing to 360°degrees. This will start at the point +1 on real axis, and move counter-clockwise around the circle until it ends up where it started, at 1 again. If the angle starts at 0 and gradually decreases until it reached −360, the point will do exactly the same thing, except rotate in a clockwise fashion.

Now we know that from the Euler equation that the complex exponential has a real and imaginary component. Try to imagine the movement of the point on the unit circle as reflected on the real axis (imagine a second point, allowed to move only on the real axis, trying to follow the first point as it moves about the circle). The movement of the second point on the real axis will equal to cos(Ω). So if we continually rotate in either direction about the unit circle, the real component will move back and forth between +1 and −1 using the motion of the cosine function. Similarly, the movement of the point on the unit circle as reflected on the imaginary axis will be similar, except instead of starting at value of +1, it will start with value 0. The pattern of motion will lag by 90°degrees. The imaginary axis movement is equal to j·sin(Ω), and the imaginary component will move back and forth between j and −j using the sine function.

This is shown in Fig. 2.6, where the dashed line represents the imaginary axis movement j·sin(Ω) and the dotted line represents the real axis movement of cos(Ω).

$X = \cos (Ω) (real axis movement)$ $X = \cos (Ω) (real axis movement)$

$Y = \sin (Ω) (imaginary axis movement)$ $Y = \sin (Ω) (imaginary axis movement)$

This gives us better way to express a complex number in polar coordinates.

Recall Z = X + jY = R angle (Ω)

As we saw before,

$X = R \cos (Ω)$ $X = R \cos (Ω)$

$Y = R \sin (Ω)$ $Y = R \sin (Ω)$

Figure 2.6 Graphing of complex exponetial.

So we can see that angle (Ω) has the same meaning as exp(jΩ). Also, for the unit circle, R = 1 by definition. So our new way to express a number in polar form using the complex exponential is:

$Z = R angle (Ω) = R e^{j Ω} (any point in complex plane)$ $Z = R angle (Ω) = R e^{j Ω} (any point in complex plane)$

$Z = angle (Ω) = e^{j Ω} (for R = 1, any point on the unit circle)$ $Z = angle (Ω) = e^{j Ω} (for R = 1, any point on the unit circle)$

This is the way you will often see it in the textbooks and industry literature.

2.7. Measuring Angles in Radians

The last curve ball in this chapter involves measuring angles in radians. You will have to get used to this, because you will see it everywhere in DSP. In our discussion, to make things more familiar, we started measuring angles in degrees, where 360°degrees describes a full circle. More commonly, the angle measurement in radians is based upon π, which is a number defined to have a value of about 3.141592 (it actually is an irrational number, with infinite number of digits, like 1/3 = 0.3333….). It takes exactly 2π radians to describe a full circle (Table 2.1).

Just like angle measurements are periodic in 360°degrees, they are also periodic in 2π radians. Using π is really no different than getting used to meters rather than using feet for measuring distances (or the reverse if you did not grow up in the United States).

We are going to see this same concept later in sampling theory, where everything tends to wraps around or behave periodically. We can visualize this as traveling either clockwise (negative rotation) or counter-clockwise (positive rotation) around the circle.

There is one more DSP convention to be aware of. The real component (we used X in discussion above) is usually called the “I” or in-phase component, and the imaginary component (we used Y in discussion above) is usually referred to as the “Q” or quadrature phase component. In many DSP algorithms, the digital signal processing must be performed simultaneously on both I and Q data streams, which we now know simply represents the signal's movement, over time, within the two dimensions of our complex number plane.

Table 2.1

Mapping Between Degrees and Radians

Angle in Degrees	Angle in Radians
0	0 π
45	π/4
90	π/2
180 = −180	π = −π
270 = −90	(3/2) π = −π/2
360 = 0	2π = 0π
−360 = 0	−2π = 0π
540 = 180	3π = π
−540 = −180 = 180	−3π = −π = π

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Table of Contents for Chapter 2. Complex Numbers and Exponentials

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