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ET4169 MICROWAVE RADAR AND REMOTE SENSING (2010-2011 Q3) 1

Laboratory report No.3 FMCW radar

Constantinescu Mihai Marius 4122992

[email protected]

Abstract

In the first measurement we acquire the radar output signal (beat signal) resulting from the

backscattering of a large metal plate for both settings of the radar bandwidth (1GHz with 23.4 ms

sweep time and 500MHz with a 11.7 ms sweep time). The recorded signal must be processed in

MATLAB to produce proper range profiles. In the second assignment we make measurements for

three reflector setups: two reflectors are separated in radar range by 20cm, 50cm and 1m. We acquire

the beat signal in these three cases with a 500MHz bandwidth and 11.7 ms sweep time.

!

1 STUDY AND DRAW THE MEASUREMENT SET-UP. THE DRAWING SHOULD SHOWTHE TARGETS, THE RADAR AND THE DISTANCES BETWEEN THEM.

Figure 1 presents the measurement set-up, containing our radar with the antenna dish and themixer (switches from 500MHz to 1 GHz and 11,7 ms / 23.4 ms sweep time) connected to thesound card of the PC for further processing. Our targets are a large metal plate for the first caseand two smaller plates positioned at 20cm, 50cm and 100cm distance one from each other, inorder to show the dependence of resolution on the bandwidth and position.

Fig. 1. Measurement Scheme

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Fig. 2. Beat Signal

2 OBTAIN FROM THE FIRST TWO MEASUREMENTS (ON THE METAL PLATE) TH E

RANGE PROFILES BY MEANS OF FF T OF THE ACQUIRED BEAT SIGNALS AND PLOT

THEM. INDICATE HOW THE FREQUENCY SCALE CAN BE CONVERTED TO DISTANCE

AND ADJUST THE PLOT TO GIVE THE DISTANCE SCALE. WHAT IS THE DISTANCE OF

THE FLAT PLATE MEASURED FROM THE RADAR? HOW DOES IT COMPARE TO THE

REAL DISTANCE?

First we aquire a beat signal, selected among the multiple beat signals (figure 2 presenst the

beat signal and figure 3 the signal after removing the triggering pulse and the DC component)and then processing is done on that signal by taking FFT of it (for better performances we areusing 8192 samples so we have to add 0 padding to the signal). Figure 4 shows the result ofapplying fft, the frequency profile and we observe our metal plate around 1200 Hz (both cases500 MHz, 1 GHz).

The range and the frequency for a FMCW radar are related with the equation (1):

R =cTsfb

2B(1)

where R represents the range, c is the speed of light, Ts sweep time (11,7 ms / 23.4 ms), fb beatfrequency and B the bandwidth (500MHz / 1 GHz). Using the above formula, the range profiles

for the plate can be plotted and are figure 5 shows the results. We can observe the distance ofthe flat metal plate measured from the radar which is 4.4 m for 1GHz and 4.3 m for 500MHz(compared with the first measurement (1GHZ), this result is more accurate towards the practicaldistance), result that is similar with the real distance measured during the experiment( 4m). Wecan also observe a second peak at 10 m representink the wall of the room.

Figure 6 compares the results of frequency and range profiles for the 2 situations (500MHz/1GHz)

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Fig. 3. Required Beat Signal (without triggering pulse)

Fig. 4. Frequency profile (magnitude)

3 INVESTIGATE THE USE OF WINDOWING ON THE BEATSIGNAL . TAKE THE PLOT

FROM 1. 2 AND ADD A PROFILE AFTER APPLYING A WINDOW, E.G. TH E HANNINGWINDOW. WHAT IS THE EFFECT?

Before doing FFT of the beat signals, we are adding a window to it, in order to smooth it.For investigating the effect of windowing on the beatsignal we are applying a Hanning and aChebysev window and the plots below present the results. Figure 7 and 8 present the Rangeprofile using Hanning window (Magnitude and Power) and figure 9 presents the range profileusing a Chebysev window while the last two figures (figures 9, 10) present a comparison ofthe results of windows (1GHz and 500 MHz case). We can observe a couple of differences(more evident in the dB plot) like peaks get larger or valleys get deeper (less riples), lessnoise (these modified signals have lower noise level). Table 1 also presents the pulse widths for

different windows.

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Fig. 5. Range profile (magnitude)

Fig. 6. Range / Frequency profiles (Power (dB)) blue=1GHz / green=500Mhz

4 DISCUSS THE COMPLETE SEQUENCE OF PROCESSING STEPS YOU HAVE AP-

PLIED IN PRODUCING THE RANGE PROFILES.As presented before we first aquired a beat signal, selected among the multiple beat signals(after removing the triggering pulse) and then processing is done on that signal by taking FFTof it (for better performances we are using 8192 samples so we have to add 0 padding to thesignal). Using equation (1), from which range can be derived using the frequency we finnalyare able to get the range profiles. So in a nutshell:

Import the beat signals from FMCW Radar Select a sweep period Add a window to the data (optional) 0 padding at the end of the beat signals

Fourier Transform of the time domain signals to frequency domain

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Fig. 7. Range profile using Hanning window (magnitude)

Fig. 8. Frequency / Range profile using Hanning window (Power(dB)) blue=1GHz / green=500Mhz

Convert beat frequency to range

Estimate distance to target

5 STUDY THE EFFECT OF RANGE RESOLUTION. MAKE THREE RANGE PROFILE

PLOTS FROM THE SECOND ASSIGNMENT, WITH TWO CURVES PER PLOT WITH AND

WITHOUT WEIGHTING FUNCTION. WHAT CAN YOU SAY OF THE RADAR RESOLUTION ?

HOW IS IT ILLUSTRATED IN THE RANGE PROFILES? WHAT IS AN IMPACT OF WINDOW-

ING ON THE RANGE RESOLUTION?

The same procedure is applied for the case of the two plates (figure 11 presents the signalsof plates at 20, 50 100 cm, in the final form without triggering pulse). Figure 12 contains the

frequency profiles (with and without (hanning) window) and figures 13, 14 and 15 present as

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Fig. 9. Frequency / Range profile using Chebyshev window (Power(dB)) blue=1GHz /

green=500Mhz

Fig. 10. Window Comparison (500MHz))

asked three range profile plots from the second assignment, with two curves per plot with andwithout weighting function.

It can be noticed from the figures that the targets were detected in all the cases, but because oflack of resolution it was difficult to detect two target plates (if the distance between two targets isonly 20 cm, the FMCW radar is not able to distinguish them, because its resolution is 30cm fromthe resolution equation (2); even the targets are 50 cm apart, they still seem to be together due tonoise or maybe the applied window is not suitable). In the last case of plate separation 100 cm-s(at frequency of 500MHz it is clearly evident that there are two target plates. We can also seethe performarces of applying a window (in this case Hanning) and we can clearly distinguishbetter our targets (separate them) compared with the initial case. We further investigate the useof a window and also aply Chebysev and Kaiser windows and then compare the results (figures

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Fig. 11. Window Comparison (1 GHZ)

Fig. 12. 2 Plates beat signal at 20, 50, 100 cm separation)

16, 17, 18) - we can say that the best results are performed by the hanning window (in the 100

cm case we can clearly distinguish between the 2 plates). For the Hanning case we can alsocompute the difference of the two peaks 4.2 m - 3.2 m and verify in this way the distance itsreally 100 cm.

Theoretically the resolution of the radar is given by equation (2):

R =c

2B(2)

where R is the resolution, c speed of light, B bandwidth so it can be clearly seen that with higherbandwidth results a better resolution (1 GHz better than 0.5GHz). The experiment results followin general the theory expectations but the difference is not that huge.

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Fig. 13. Frequency profile (without and with Hanning window)

Fig. 14. Range profile (with and without weighting function)

6 STUDY THE SYSTEM IMPULSE RESPONSE IN RANGE BASED ON THE PLATE MEA-

SUREMENT OF ASSIGNMENT 1.2. WHAT CAN BE SAID OF THE IMPULSE WIDTH INCOMPARISON TO THE RANGE RESOLUTION AS DETERMINED IN 1.5? WHAT SHOULD

THE RELATION BE IN THEORY? DO THEORY AND MEASUREMENT FIT? WHAT IS THE

EFFECT OF THE BANDWIDTH ON THE RANGE RESOLUTION ?

As presented at previous point in theory the resolution of the radar is given by equation (2):

R =c

2B(3)

where R is the resolution, c speed of light, B bandwidth so it can be clearly seen that withhigher bandwidth results a better resolution (1 GHz better than 0.5GHz - 15 cm and 30 cm).

The experiment results follow in general the theory expectations but the difference is not that

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huge. From our results it can be seen that for higher frequency the pulse width reduces, which

should be the case ideally because with high frequency the resolution increases and thus thepulse width should reduce. Table 1 presents the pulse width (3 Db beamwidth) for all of thewindows in the 1GHz and 500MHz cases. Compared to 0.15 m and 0.3 m, the experimentalresolutions are bigger than the teoretical value. The reason may be the utilization of windows,that will widen the resolution.

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Fig. 17. Window comparison 100cm


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Fig. 20. Table 1: Pulse Width

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