Complex numbers

Fourier transform The Fourier transform of a continuous-time signal may be defined as The discrete version of this is Both have a j term so we need a basic understanding of complex numbers

Complex numbers There are 2 parts to a complex number, a real part and an imaginary part z=a±jb

Complex number application

But what about this one?

Complex arithmetic Add: (a+jb) + (c+jd) = (a+c) + j(b+d) e.g. (4+j5) + (3-j2) = (7+j3) Subtract: (a+jb) – (c+jd) = (a-c) + j(b-d) e.g. (4+j7) – (2-j5) = (2+j12) Multiply: (a+jb)(c+jd) = ac+jad+jbc+j2bd e.g. (3+j4)(2+j5) = 6+j15+j8+j220 = -14+j23 Note that all results are complex

Using complex arithmetic, check that the previous quadratic has the solutions:

Complex conjugate Complex conjugate: flip the sign of j e.g. the complex conjugate of (5+j8) is (5-j8) Multiplication of a complex number with its conjugate produces a real number e.g. (5+j8)(5-j8) = 25-j40+j40-j264 = 25+64=89 Now consider division of complex numbers But what about

Complex division Division: make the denominator real by multiplying top and bottom by the denominator’s complex conjugate

Complex numbers as vectors Complex numbers can not be enumerated but they can be represented diagrammatically A vector (line with magnitude and direction) of a number pointing at an angle of zero can be represented as a line on the +ve x axis Multiply the vector by -1 and it points the other way i.e. a 180° shift As -1 = j2 then j lies between them i.e. a 90° shift

Complex plane or argand diagram j -c+jd Imaginary a+jb j2= -1 Real g-jh -e-jf j3=-j

Polar form of a complex number The number may be represented by its vector magnitude and angle

Or using trig for X and Y

Polar manipulation It is easy to multiply and divide complex numbers in this form even if using degrees

Remember the series form?

Summary It is easy to add and subtract in Cartesian form It is easy to multiply and divide in polar form The exponential form is useful when dealing with sine and cosine waveforms

One final note, using even-odd trig identities (or looking at waveforms), because the Fourier transform uses e-θ: