DISCRETE-SIGNAL ANALYSIS AND DESIGN- P11 doc

A two-tone input signal of adjustable peak amplitude will be processed by a circuit that has a certain transfer characteristic which is similar to the Child-Langmuir equation [Seely, 195

to return to the time domain with the conÞdence that these two equations, especially in discrete sequences, do not require linearity or superposition

We will use this idea frequently in this book

A two-tone input signal of adjustable peak amplitude will be processed

by a circuit that has a certain transfer characteristic which is similar to the Child-Langmuir equation [Seely, 1956, pp 24 and 28, Eq (2-14)] as derived in the early 1920s from Poisson’s equation for the electric Þeld in space-charge-limited diodes and also many common triode vacuum tubes:

Iout(n) ≈ KV g (n) 1.5 (2-3)

The input (base-to-emitter or grid-to-cathode) two-tone signal at

fre-quencies f1 and f2 is

V g (n) = V s

 cos



2πn

N f1

 + cos



2πn

N f2



+ Vdc (2-4)

V s is the peak amplitude (1/2 of pk-to-pk) of each of the two input

sig-nals V dc is a bias voltage that determines the dc operating point for the

particular device This and a reasonable V s value are found from

Hand-book V-I curves (the maximum peak-to-peak signal is four times V s) The peak-to-peak ac signal should not drive the device into cutoff or saturation

or into an excessively nonlinear region Figure 2-5 is a typical approxi-mate spectrum for the two-tone output signal The input frequencies are

f1and f2, and the various intermodulation products are labeled Adjusting

V s and Vdc for a constant value of peak desired per-tone output shows

how distortion products vary Note also the addition of 70 dB to the ver-tical axis This brings up the levels of weak products so that they show prominently above the zero dB baseline (we are usually interested in the

dB differences in the spectrum lines) Note also that the vertical scale for the spectrum values is the magnitude in dB because the actual values are

in many cases complex, and we want the magnitude and not just the real part (we neglect for now the phase angles)

Note that in this example we let Mathcad calculate Vsig(n) 1.5 directly (the easy way), not by using discrete math (the hard way), just as we do with the exp(·), sin(·), cos(·) and the other functions We are especially

−8

−6

−4

−12

20

40

60

80

100

k

f1 f2

2f2−f1

2f1−f2

f1+f2

2f1 2f2

f2−f1

DC

2f1 +f2

2f2 +f1

3f1 3f2

Vsig(n)

Vin(n))

Vdc

N := 64

Vdc : = −8

Vout(n) : = Vsig(n)1.5

Vsig(n) : = Vdc + 1.5⋅

n := 0, 1 63

k := 0, 1 63 Vin(n) : = 25⋅n

cos 2 ⋅p⋅N n⋅4 + cos 2 ⋅p⋅N n ⋅5 1

N

Fip(k) := ⋅ Vout(n)⋅exp −j⋅2⋅p⋅n

N ⋅k

∑N−1

n = 0

Figure 2-5 Intermodulation measurements on an ampliÞer circuit.

interested in discrete sequences and discrete ways to process them, but

we also use Mathcad’s numerical abilities when it is sensible to do so In embedded signal-processing circuitry, machine language subroutines do all of this “grunt” work In this book we let Mathcad do it in an elegant fashion

400

Iout(n)

5

10

15

20

25

30

35

40

n

k

f1 f2

f2–f1

DC

2f1 2f2

f2+f1

1 N

Fip (k) : = ⋅ Iout (n) ⋅ exp −j ⋅ k ⋅2⋅π ⋅ n N

Vg (n) := Vb ⋅⎛⎛ cos 2 ⎛⎛ ⋅π⋅N n ⋅12 ⎛ ⎛ + cos ⎛⎛ 2 ⋅π⋅N n ⋅15 ⎛ ⎛ ⎛ ⎛ Iout (n) := Vg (n)2

N := 128 n := 0,.01 N − 1 k : = 0, 1 N − 1 Vb := 10

n = 0

Figure 2-6 Square-law ampliÞer, mixer, and frequency doubler.

In Fig 2-6 the exponent in Eq (2-3) is changed from 1.5 to 2.0 This

is the well-known square-law device that is widely used as a modulator

(mixer) or frequency doubler [Terman, 1943, p 565] Note the absence

of dc bias (optional) We see that frequencies f1 and f2 have disappeared

from the output, the sum and difference of f1 and f2 are prominent and

the second harmonics of f1 and f2 are strong also Higher-order IMD products have also vanished

The nonlinearity in Eq (2-3) can be customized for a wide variety

of devices, based on their transfer characteristics, to explore ac circuit

performance For example, Eq (2-3) can be in the form of an N -point

Handbook lookup table for transistor or tube V–I curves Pick 16 equally

spaced values of Vin(n) for n= 0 to 15 and estimate as accurately as

pos-sible the corresponding values of Iout(n) Then get the positive frequency

spectrum for low-order (2nd or 3rd) intermodulation products Nonlinear circuit simulation programs such as Multisim can explore these problems

in greater detail, using the correct dc and RCL components and accurate

slightly nonlinear device models

Example 2-2: Analysis of the Ramp Function

This chapter concludes with an analysis of the “ramp” function in Fig 2-7a It is shown in many references such as [Zwillinger, 1996, p 49] Its Fourier series equation in the Reference is

f (x)= 12 −

∞



k=1

1

πk sin

2πxk

where x is the distance along the x-axis The term 1/2 is the average height of the ramp, and x always lies between 0 and + L The sine-wave harmonics (k) extend from 1 to ∞, each with peak amplitude 1/πk For each value of (k), f (x) creates (k) sine waves within the length L The

minus sign means that the sine waves are inverted with respect to the

x axis.

In Fig 2-7 the discrete form of the ramp is shown very simply in part

(a) as x(n) = n/N from 0 ≤ n ≤ N −1 We then apply the DFT (Eq 1-2)

to get the two-sided phasor spectrum X (k) from 0 to N − 1 in part (b) The following comments help to interpret part (b):

• The real part has the value X (0) = 0.5 at k = 0, the dc value of the

ramp The actual value shown is 0.484, not 0.5, because the value 0.5 is approached only when the number of samples is very large

For N= 29= 512, X (0) is 0.499 This is an example of the

approxi-mations in discrete signal processing For very accurate answers that

we probably will never need, we could use 212and the FFT

• The real part of X (k) from k = 1 to N − 1 is negative, and part (c)

shows that the sum of these real parts is −0.484, the negative of

X (0) In part (c) the average of the real part from 0 to N− 1 is zero, which is correct for a spectrum of sine waves with no dc bias

0 5 10 15 20 25 30

0

0.5

1

x(n)

n

0 5 10 15

(c) (b)

(a)

20 25 30

−0.2 0 0.2

0.4 + 0.484

k

x(n) := n

N

N := 2 5 n : = 0, 1 N − 1 k : = 0, 1 N − 1

1

N X(k) : = ⋅ x(n)⋅exp −j⋅2⋅π⋅n

N ⋅k

Re(X(k))= −0.484 Im(X(k))= 10 −15

Im(X(k))

0

Re(X(k))

∑N−1

n = 0

∑N−1

Figure 2-7 Analysis of a ramp function.

• The imaginary part of the plot is positive-going in the Þrst half,

negative-going in the second half, and zero at N /2 Referring to

Fig 2-2, this arrangement of polarity agrees with the− sin diagram,

as it should

• Applying the “Mathcad X-Y Trace” tool to the imaginary part of the spectrum plot in part (b), we Þnd that the two sides are odd-symmetric

(Hermitian) about N /2 For this reason, the relative phase is zero for

these sine waves (they all begin and end at zero phase)