# Clarification for “Digital Signal Processing” by Steven W. Smith related to complex representation of sinusoids

I have a question related to complex number representation of sinusoids from a book by Steven W. Smith "Digital Signal Processing". The book is made freely available by the author on dspguide.com. The part I have question about is available here, pp.559-563.

Excerpt from the book (p. 560):

A sinusoid is represented by a complex number in rectangular form as follows:

$$A \cos(\omega t) + B \sin(\omega t) \rightleftharpoons a + jb$$

where $$a \rightleftharpoons A$$ and $$b \rightleftharpoons -B$$.

A sinusoid is represented by a complex number in polar form as follows:

$$M \cos(\omega t + \phi) \rightleftharpoons M e^{j\theta}$$

where $$\theta \rightleftharpoons -\phi$$.

Why change the sign of the imaginary part and phase angle? This is to make the substitution appear in the same form as the complex Fourier transform described in the next chapter. The substitution techniques of this chapter gain nothing from this sign change, but it is almost always done to keep things consistent with the more advanced methods.

What I am confused about is that these two representations are not equivalent. Here is an example following the rules described above for rectangular and polar forms:

$$\cos(\omega t) + \sin(\omega t) \rightleftharpoons 1 - j = \sqrt{2} e^{-j\frac{\pi}{4}}$$

$$\cos(\omega t) + \sin(\omega t) = \sqrt{2} \cos(\omega t - \frac{\pi}{4}) \rightleftharpoons \sqrt{2} e^{j\frac{\pi}{4}} = 1 + j$$

The rules for rectangular and polar form produce phase angles of opposite sign.

Author gives few more examples on p. 561 (first paragraph in "Complex Represenation of Systems") that show clear discrepancy between rectangular and polar forms considering phase angle.

I am aware that we can take either rectangular or polar form to "be correct", the sign is a convention anyway. But how can both of them be used at the same time?

Phasors are useful for the analysis of (real-valued) linear time-invariant (LTI) systems. A phasor is a complex number

$$C=|C|e^{j\phi}=A+jB\tag{1}$$

which represents a sinusoidal signal:

$$x(t)=\textrm{Re}\big\{Ce^{j\omega t}\big\}=A\cos(\omega t)-B\sin(\omega t)=|C|\cos(\omega t+\phi)\tag{2}$$

That's already the end of the story. Eq. $$(2)$$ is the generally accepted definition of a phasor. The definition in the book is either a mistake or an idiosyncrasy making things unnecessarily confusing.

• As the excerpt from the book says, the sign conversion is there to match it with later chapters. I was confused about rectangular and polar form not matching, and thought that I might be missing some "bigger picture" here. I will try to contact the author and see what he says about it. – Marko Jan 26 at 10:48
• @Marko: I'm not sure what is meant by "matching with later chapters", but it seems bizarre to define a phasor differently depending on the use of polar or rectangular representation. I haven't seen such a definition anywhere else. – Matt L. Jan 26 at 10:51
• It is related to complex Fourier transform where this sign inversion exists naturally in the equation, and is often overlooked by engineers. My assumption is that author just wanted to stress negative sign with imaginary part. He even mentions this in the book: The substitution techniques of this chapter gain nothing from this sign change, but it is almost always done to keep things consistent with the more advanced methods. The sign inversion is okay if you are aware what you are doing, but rectangular and polar form must result in the same complex number. I will post here what author says. – Marko Jan 26 at 10:56