Adding a waveform on top of an LED driver signal
Your choice of common-cathode red-green-blue LEDs means that an analog LED driver should drive LED current into the open LED anode. The common LED cathode would go to 0V (GND).
But most signal sources provide output relative to ground. This makes driving a cathode-grounded LED with current a bit more complex. Your AD9833 signal source is of this type. Many microphones are too.
Your LED suggests maximum current of 30mA. This means that with no AC signal, it should idle at half-current, at 15mA. Then it can swing down toward zero mA, and up toward 30mA with signal current. The LED would appear constantly bright, because your eye cannot see rapid signal swings at frequencies above 30 Hz. Average current remains at about 15 mA.
An analog driver, driving a cathode-grounded LED might look like the two-transistor circuit (below). Resistor bias is chosen so that most any LED would have roughly 15mA flowing with no input signal, provided the DC supply voltage is +5V. Input voltage to C1 would be about 0.65V peak-to-peak AC signal.

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Driving common-cathode LED array is simpler with a PWM signal source. You only have to calculate LED current for two cases:

LED OFF: 0 mA current
LED ON: 30 mA current

A transistor saturated switch toggles between these two cases, driven from a digital signal source. An Arduino with a Vdd supply of +5V would toggle a digital output pin between two voltages: 0V &amp; +5V. This signal would drive a switching transistor ON or OFF.
Current from the Arduino pin is small (1.4 mA), limited by a resistor (R1, below). Another resistor (R2) is chosen to limit LED current to 30 mA, from a +5v DC supply. The transistor has ample current gain and is actually over-driven so that it acts as an ON/OFF switch. Note that the 30 mA current through transistor and LED does not flow from the microcontroller, but does flow back to the +5V supply. Many PWM outputs could drive such a switch before microcontroller current limits are exceeded.



simulate this circuit &ndash; Schematic created using CircuitLab
 A PWM signal of 50% duty cycle would drive the LED ON to 30 mA half the time, and drive the LED OFF to 0 mA half the time. The average is 15 mA.
Converting an analog waveform input to PWM signal is done with software inside Arduino, which uses its analog-to-digital converter (ADC) to provide to its PWM module a digital input. The PWM module would be set for a frequency above the audio range for your application (above 20 kHz).
Most of these microcontroller ADCs use +Vdd as a reference voltage, so that input signal ranging between 0 -to- +5V gives an output value between 0 -to- 256 for an 8-bit ADC, for example.
A program might read ADC 20,000 times per second, convert this number to a range compatible with PWM range, and write to the PWM duty cycle register.

Driving multiple LEDs with one transistor switch:



simulate this circuit
The MOSfet would be selected from data sheet to have less than about one ohm \$RDS_{on}\$. A &quot;logic level&quot; MOSfet would be preferred, especially if the microcontroller was run at a supply voltage lower than +5V. The MOSfet should not run hot with only 90mA being switched on/off.
Since the switched 90mA current is reasonably high, a DC supply bypass capacitor should be placed near the resistor common point, and MOSfet source (which is grounded). This should reduce radiated EMI interference.
When you drive multiple LEDs with ONE PWM switch like this, each LED is modulated with the SAME PWM signal. If you want each LED to have individually-controlled modulation, each LED would require its own PWM channel, and its own MOSfet. Some microcontrollers have multiple PWMs on board.