frequency sweep simulation challenge

[maxidivine]

Hello,

 I have been trying to simulate two speed signals with the following specification.

Signal 1:

Maximum amplitude:    5V

Minimum amplitude:     12V

Form:                                    Sinusoidal

OP Impedance:                 ~100 Ohm

Switchable:                         8Hz or 800Hz

 

Signal 2:

Maximum amplitude:    5V

Minimum amplitude:     12V

Form:                                    Sinusoidal

OP Impedance:                 ~100 Ohm

Adjustable over:              8Hz to 1000Hz

Soft switchable sweep: 800 Hz rising @ 7% per second

                                                800 Hz rising @ 9% per second

                                                800Hz rising @ 11% per second.

I have implemented the first part. For the second one,  I have used the formula for the sweep, taking the 7% per second as an example.

From v = u + at where v is freqency, and then integrating:

800Hz to 1.07*800:                          800 + 56t

856 Hz to 1.07*856;                         856 + 59.2t

915.92Hz to 980.0344;                915.92 + 980.0344t

sin((800* t) + ((56*t*t)/2))+sin((856* t) + ((59.2*t*t)/2))+sin((915.92* t) + ((980.0344*t*t)/2))

 I have tried simulating two portions of this sine wave. But I cannot implement the whole thing on a single waveform.

Please help. I have attached the Vis for 1 and a trial for the second one.

 

Thanks.

[Ashish_Naik_NIUK]

Hey there,

Are you looking at doing a linear sweep? A really good example LabVIEW code for showing a broad range of analog signals, including a linear sweep is shown http://zone.ni.com/devzone/cda/epd/p/id/4623#0requirements

From what it sounds like you are wanting to do a Chirp signal. A chirp signal is a signal which continuously increases in frequency (or down). This is explained in the KnowledgeBase article KB 4HHCCOD2

Please let us know how your getting on.

Kind regards,

[maxidivine]

Thanks a lot for the wonderful response. I intend trying it out using the formula:

f = sin(2*pi[(f+(rate*t)]t ) i.e from f = sin(2*pi*f*t)

f = initial frequeny, in my case, 800Hz

   rate = rate of sweep increase with options of 7%, 9% and 11% in my case.

t= time in seconds.

I hope to implement it using the formula waveform vi.

I would get back with the VI later in the day.

Thanks a lot.

[Ashish_Naik_NIUK]

Thanks for letting me know. Please give us a shout with how you get on!

Ash

[maxidivine]

Hi,

I have been able to come up with this VI to get a sweep for a 7% increasing frequency. Please have a look at the VI and let me know what you think.

I have attached the VI here.

from sin (2*pi*f*t)

sin(2*pi*f(t)*t)

f(t) = f0(1+at)   where f0 is initial frequency and a is the rate of increase in this case, 7%

sin(2*pi*f0(1+at)*t)

[Ashish_Naik_NIUK]

Hi maxidivine,

I am glad that you have got things working! I have if you dont mind edited the vi to give you another control to be able to change the % of the rise per second, as I noticed this was a requirement of yours.

Please let me know what you think.

Ash

[maxidivine]

Thanks so much. It is amazing just how powerful LabVIEW is. I have resumed work on a datalogging project and things are going well. I would not hesitate to seek for assistance from this very useful and encouraging forum.

Thanks.