如何发出所有正在等待的go例程的信号？

The problem I have is to "signal all go routines that are waiting". Some go routines may be performing other tasks (simulated by time.Sleep in code below), before waiting for the signal. Others go routines may wait for the signal immediately after starting. In some case there may not be any go routines that are waiting (not simulated in the code below)

I came with the following approach:

package main

import (
    "fmt"
    "sync"
    "time"
)

func main() {

    wg := sync.WaitGroup{}
    wg.Add(3)
    cond := sync.NewCond(&sync.Mutex{})
    signalled := false

    producer := func() {
        time.Sleep(1)
        cond.L.Lock()
        signalled = true
        cond.Broadcast()
        cond.L.Unlock()
        wg.Done()
    }
    go producer()

    go func() {
        time.Sleep(3 * time.Second)
        cond.L.Lock()
        for !signalled {
            cond.Wait()
        }
        cond.L.Unlock()
        wg.Done()
    }()

    go func() {
        cond.L.Lock()
        for !signalled {
            cond.Wait()
        }
        cond.L.Unlock()
        wg.Done()
    }()

    wg.Wait()
    fmt.Println("done once")

    reset := func() {
        cond.L.Lock()
        signalled = false
        cond.L.Unlock()
    }
    // a reset function
    reset()

    wg.Add(2)
    go producer()
    go func() {
        cond.L.Lock()
        for !signalled {
            cond.Wait()
        }
        cond.L.Unlock()
        wg.Done()
    }()
    wg.Wait()
    fmt.Println("done twice")
}

https://play.golang.org/p/uNmyofH2hs

I can use a channel and close the channel as a signal. But this means that in case there is a need for "reset" (where in the "signal" can be triggered again), the channel would have to be created again.

Is there a better (concise) way to achieve this?